mirror of
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1671 lines
67 KiB
Markdown
1671 lines
67 KiB
Markdown
---
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comments: true
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---
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# 13.3 Subset-Sum Problem
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## 13.3.1 Without Duplicate Elements
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!!! question
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Given a positive integer array `nums` and a target positive integer `target`, find all possible combinations where the sum of elements in the combination equals `target`. The given array has no duplicate elements, and each element can be selected multiple times. Return these combinations in list form, where the list should not contain duplicate combinations.
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For example, given the set $\{3, 4, 5\}$ and target integer $9$, the solutions are $\{3, 3, 3\}, \{4, 5\}$. Note the following two points:
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- Elements in the input set can be selected repeatedly without limit.
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- Subsets do not distinguish element order; for example, $\{4, 5\}$ and $\{5, 4\}$ are the same subset.
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### 1. Reference to Full Permutation Solution
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Similar to the full permutation problem, we can imagine the process of generating subsets as a series of choices, and update the "sum of elements" in real-time during the selection process. When the sum equals `target`, we record the subset to the result list.
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Unlike the full permutation problem, **elements in this problem's set can be selected unlimited times**, so we do not need to use a `selected` boolean list to track whether an element has been selected. We can make minor modifications to the full permutation code and initially obtain the solution:
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=== "Python"
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```python title="subset_sum_i_naive.py"
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def backtrack(
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state: list[int],
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target: int,
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total: int,
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choices: list[int],
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res: list[list[int]],
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):
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"""Backtracking algorithm: Subset sum I"""
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# When the subset sum equals target, record the solution
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if total == target:
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res.append(list(state))
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return
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# Traverse all choices
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for i in range(len(choices)):
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# Pruning: if the subset sum exceeds target, skip this choice
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if total + choices[i] > target:
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continue
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# Attempt: make choice, update element sum total
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state.append(choices[i])
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# Proceed to the next round of selection
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backtrack(state, target, total + choices[i], choices, res)
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# Backtrack: undo choice, restore to previous state
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state.pop()
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def subset_sum_i_naive(nums: list[int], target: int) -> list[list[int]]:
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"""Solve subset sum I (including duplicate subsets)"""
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state = [] # State (subset)
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total = 0 # Subset sum
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res = [] # Result list (subset list)
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backtrack(state, target, total, nums, res)
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return res
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```
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=== "C++"
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```cpp title="subset_sum_i_naive.cpp"
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/* Backtracking algorithm: Subset sum I */
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void backtrack(vector<int> &state, int target, int total, vector<int> &choices, vector<vector<int>> &res) {
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// When the subset sum equals target, record the solution
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if (total == target) {
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res.push_back(state);
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return;
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}
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// Traverse all choices
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for (size_t i = 0; i < choices.size(); i++) {
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// Pruning: if the subset sum exceeds target, skip this choice
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if (total + choices[i] > target) {
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continue;
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}
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// Attempt: make choice, update element sum total
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state.push_back(choices[i]);
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// Proceed to the next round of selection
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backtrack(state, target, total + choices[i], choices, res);
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// Backtrack: undo choice, restore to previous state
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state.pop_back();
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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vector<vector<int>> subsetSumINaive(vector<int> &nums, int target) {
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vector<int> state; // State (subset)
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int total = 0; // Subset sum
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vector<vector<int>> res; // Result list (subset list)
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backtrack(state, target, total, nums, res);
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return res;
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}
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```
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=== "Java"
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```java title="subset_sum_i_naive.java"
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/* Backtracking algorithm: Subset sum I */
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void backtrack(List<Integer> state, int target, int total, int[] choices, List<List<Integer>> res) {
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// When the subset sum equals target, record the solution
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if (total == target) {
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res.add(new ArrayList<>(state));
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return;
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}
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// Traverse all choices
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for (int i = 0; i < choices.length; i++) {
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// Pruning: if the subset sum exceeds target, skip this choice
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if (total + choices[i] > target) {
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continue;
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}
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// Attempt: make choice, update element sum total
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state.add(choices[i]);
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// Proceed to the next round of selection
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backtrack(state, target, total + choices[i], choices, res);
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// Backtrack: undo choice, restore to previous state
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state.remove(state.size() - 1);
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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List<List<Integer>> subsetSumINaive(int[] nums, int target) {
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List<Integer> state = new ArrayList<>(); // State (subset)
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int total = 0; // Subset sum
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List<List<Integer>> res = new ArrayList<>(); // Result list (subset list)
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backtrack(state, target, total, nums, res);
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return res;
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}
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```
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=== "C#"
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```csharp title="subset_sum_i_naive.cs"
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/* Backtracking algorithm: Subset sum I */
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void Backtrack(List<int> state, int target, int total, int[] choices, List<List<int>> res) {
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// When the subset sum equals target, record the solution
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if (total == target) {
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res.Add(new List<int>(state));
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return;
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}
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// Traverse all choices
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for (int i = 0; i < choices.Length; i++) {
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// Pruning: if the subset sum exceeds target, skip this choice
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if (total + choices[i] > target) {
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continue;
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}
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// Attempt: make choice, update element sum total
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state.Add(choices[i]);
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// Proceed to the next round of selection
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Backtrack(state, target, total + choices[i], choices, res);
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// Backtrack: undo choice, restore to previous state
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state.RemoveAt(state.Count - 1);
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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List<List<int>> SubsetSumINaive(int[] nums, int target) {
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List<int> state = []; // State (subset)
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int total = 0; // Subset sum
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List<List<int>> res = []; // Result list (subset list)
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Backtrack(state, target, total, nums, res);
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return res;
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}
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```
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=== "Go"
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```go title="subset_sum_i_naive.go"
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/* Backtracking algorithm: Subset sum I */
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func backtrackSubsetSumINaive(total, target int, state, choices *[]int, res *[][]int) {
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// When the subset sum equals target, record the solution
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if target == total {
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newState := append([]int{}, *state...)
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*res = append(*res, newState)
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return
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}
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// Traverse all choices
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for i := 0; i < len(*choices); i++ {
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// Pruning: if the subset sum exceeds target, skip this choice
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if total+(*choices)[i] > target {
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continue
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}
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// Attempt: make choice, update element sum total
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*state = append(*state, (*choices)[i])
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// Proceed to the next round of selection
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backtrackSubsetSumINaive(total+(*choices)[i], target, state, choices, res)
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// Backtrack: undo choice, restore to previous state
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*state = (*state)[:len(*state)-1]
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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func subsetSumINaive(nums []int, target int) [][]int {
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state := make([]int, 0) // State (subset)
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total := 0 // Subset sum
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res := make([][]int, 0) // Result list (subset list)
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backtrackSubsetSumINaive(total, target, &state, &nums, &res)
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return res
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}
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```
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=== "Swift"
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```swift title="subset_sum_i_naive.swift"
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/* Backtracking algorithm: Subset sum I */
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func backtrack(state: inout [Int], target: Int, total: Int, choices: [Int], res: inout [[Int]]) {
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// When the subset sum equals target, record the solution
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if total == target {
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res.append(state)
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return
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}
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// Traverse all choices
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for i in choices.indices {
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// Pruning: if the subset sum exceeds target, skip this choice
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if total + choices[i] > target {
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continue
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}
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// Attempt: make choice, update element sum total
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state.append(choices[i])
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// Proceed to the next round of selection
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backtrack(state: &state, target: target, total: total + choices[i], choices: choices, res: &res)
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// Backtrack: undo choice, restore to previous state
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state.removeLast()
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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func subsetSumINaive(nums: [Int], target: Int) -> [[Int]] {
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var state: [Int] = [] // State (subset)
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let total = 0 // Subset sum
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var res: [[Int]] = [] // Result list (subset list)
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backtrack(state: &state, target: target, total: total, choices: nums, res: &res)
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return res
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}
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```
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=== "JS"
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```javascript title="subset_sum_i_naive.js"
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/* Backtracking algorithm: Subset sum I */
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function backtrack(state, target, total, choices, res) {
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// When the subset sum equals target, record the solution
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if (total === target) {
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res.push([...state]);
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return;
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}
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// Traverse all choices
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for (let i = 0; i < choices.length; i++) {
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// Pruning: if the subset sum exceeds target, skip this choice
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if (total + choices[i] > target) {
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continue;
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}
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// Attempt: make choice, update element sum total
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state.push(choices[i]);
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// Proceed to the next round of selection
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backtrack(state, target, total + choices[i], choices, res);
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// Backtrack: undo choice, restore to previous state
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state.pop();
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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function subsetSumINaive(nums, target) {
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const state = []; // State (subset)
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const total = 0; // Subset sum
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const res = []; // Result list (subset list)
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backtrack(state, target, total, nums, res);
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return res;
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}
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```
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=== "TS"
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```typescript title="subset_sum_i_naive.ts"
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/* Backtracking algorithm: Subset sum I */
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function backtrack(
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state: number[],
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target: number,
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total: number,
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choices: number[],
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res: number[][]
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): void {
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// When the subset sum equals target, record the solution
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if (total === target) {
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res.push([...state]);
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return;
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}
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// Traverse all choices
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for (let i = 0; i < choices.length; i++) {
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// Pruning: if the subset sum exceeds target, skip this choice
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if (total + choices[i] > target) {
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continue;
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}
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// Attempt: make choice, update element sum total
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state.push(choices[i]);
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// Proceed to the next round of selection
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backtrack(state, target, total + choices[i], choices, res);
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// Backtrack: undo choice, restore to previous state
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state.pop();
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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function subsetSumINaive(nums: number[], target: number): number[][] {
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const state = []; // State (subset)
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const total = 0; // Subset sum
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const res = []; // Result list (subset list)
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backtrack(state, target, total, nums, res);
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return res;
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}
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```
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=== "Dart"
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```dart title="subset_sum_i_naive.dart"
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/* Backtracking algorithm: Subset sum I */
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void backtrack(
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List<int> state,
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int target,
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int total,
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List<int> choices,
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List<List<int>> res,
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) {
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// When the subset sum equals target, record the solution
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if (total == target) {
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res.add(List.from(state));
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return;
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}
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// Traverse all choices
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for (int i = 0; i < choices.length; i++) {
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// Pruning: if the subset sum exceeds target, skip this choice
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if (total + choices[i] > target) {
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continue;
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}
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// Attempt: make choice, update element sum total
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state.add(choices[i]);
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// Proceed to the next round of selection
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backtrack(state, target, total + choices[i], choices, res);
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// Backtrack: undo choice, restore to previous state
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state.removeLast();
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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List<List<int>> subsetSumINaive(List<int> nums, int target) {
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List<int> state = []; // State (subset)
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int total = 0; // Sum of elements
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List<List<int>> res = []; // Result list (subset list)
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backtrack(state, target, total, nums, res);
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return res;
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}
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```
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=== "Rust"
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```rust title="subset_sum_i_naive.rs"
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/* Backtracking algorithm: Subset sum I */
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fn backtrack(
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state: &mut Vec<i32>,
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target: i32,
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total: i32,
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choices: &[i32],
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res: &mut Vec<Vec<i32>>,
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) {
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// When the subset sum equals target, record the solution
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if total == target {
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res.push(state.clone());
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return;
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}
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// Traverse all choices
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for i in 0..choices.len() {
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// Pruning: if the subset sum exceeds target, skip this choice
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if total + choices[i] > target {
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continue;
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}
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// Attempt: make choice, update element sum total
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state.push(choices[i]);
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// Proceed to the next round of selection
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backtrack(state, target, total + choices[i], choices, res);
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// Backtrack: undo choice, restore to previous state
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state.pop();
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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fn subset_sum_i_naive(nums: &[i32], target: i32) -> Vec<Vec<i32>> {
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let mut state = Vec::new(); // State (subset)
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let total = 0; // Subset sum
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let mut res = Vec::new(); // Result list (subset list)
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backtrack(&mut state, target, total, nums, &mut res);
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res
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}
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```
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=== "C"
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```c title="subset_sum_i_naive.c"
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/* Backtracking algorithm: Subset sum I */
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void backtrack(int target, int total, int *choices, int choicesSize) {
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// When the subset sum equals target, record the solution
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if (total == target) {
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for (int i = 0; i < stateSize; i++) {
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res[resSize][i] = state[i];
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}
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resColSizes[resSize++] = stateSize;
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return;
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}
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// Traverse all choices
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for (int i = 0; i < choicesSize; i++) {
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// Pruning: if the subset sum exceeds target, skip this choice
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if (total + choices[i] > target) {
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continue;
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}
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// Attempt: make choice, update element sum total
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state[stateSize++] = choices[i];
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// Proceed to the next round of selection
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backtrack(target, total + choices[i], choices, choicesSize);
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// Backtrack: undo choice, restore to previous state
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stateSize--;
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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void subsetSumINaive(int *nums, int numsSize, int target) {
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resSize = 0; // Initialize solution count to 0
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backtrack(target, 0, nums, numsSize);
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}
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```
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=== "Kotlin"
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```kotlin title="subset_sum_i_naive.kt"
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/* Backtracking algorithm: Subset sum I */
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fun backtrack(
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state: MutableList<Int>,
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target: Int,
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total: Int,
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choices: IntArray,
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res: MutableList<MutableList<Int>?>
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) {
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// When the subset sum equals target, record the solution
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if (total == target) {
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res.add(state.toMutableList())
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return
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}
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// Traverse all choices
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for (i in choices.indices) {
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// Pruning: if the subset sum exceeds target, skip this choice
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if (total + choices[i] > target) {
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continue
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}
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// Attempt: make choice, update element sum total
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state.add(choices[i])
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// Proceed to the next round of selection
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backtrack(state, target, total + choices[i], choices, res)
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// Backtrack: undo choice, restore to previous state
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state.removeAt(state.size - 1)
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}
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}
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/* Solve subset sum I (including duplicate subsets) */
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fun subsetSumINaive(nums: IntArray, target: Int): MutableList<MutableList<Int>?> {
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val state = mutableListOf<Int>() // State (subset)
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val total = 0 // Subset sum
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val res = mutableListOf<MutableList<Int>?>() // Result list (subset list)
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backtrack(state, target, total, nums, res)
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return res
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}
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```
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=== "Ruby"
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```ruby title="subset_sum_i_naive.rb"
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### Backtracking: subset sum I ###
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def backtrack(state, target, total, choices, res)
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# When the subset sum equals target, record the solution
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if total == target
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res << state.dup
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return
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end
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# Traverse all choices
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for i in 0...choices.length
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# Pruning: if the subset sum exceeds target, skip this choice
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next if total + choices[i] > target
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# Attempt: make choice, update element sum total
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state << choices[i]
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# Proceed to the next round of selection
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backtrack(state, target, total + choices[i], choices, res)
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# Backtrack: undo choice, restore to previous state
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state.pop
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end
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end
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### Solve subset sum I (with duplicate subsets) ###
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def subset_sum_i_naive(nums, target)
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state = [] # State (subset)
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total = 0 # Subset sum
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res = [] # Result list (subset list)
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backtrack(state, target, total, nums, res)
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res
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end
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```
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When we input array $[3, 4, 5]$ and target element $9$ to the above code, the output is $[3, 3, 3], [4, 5], [5, 4]$. **Although we successfully find all subsets that sum to $9$, there are duplicate subsets $[4, 5]$ and $[5, 4]$**.
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This is because the search process distinguishes the order of selections, but subsets do not distinguish selection order. As shown in Figure 13-10, selecting 4 first and then 5 versus selecting 5 first and then 4 are different branches, but they correspond to the same subset.
|
|
|
|
{ class="animation-figure" }
|
|
|
|
<p align="center"> Figure 13-10 Subset search and boundary pruning </p>
|
|
|
|
To eliminate duplicate subsets, **one straightforward idea is to deduplicate the result list**. However, this approach is very inefficient for two reasons:
|
|
|
|
- When there are many array elements, especially when `target` is large, the search process generates many duplicate subsets.
|
|
- Comparing subsets (arrays) is very time-consuming, requiring sorting the arrays first, then comparing each element in them.
|
|
|
|
### 2. Pruning Duplicate Subsets
|
|
|
|
**We consider deduplication through pruning during the search process**. Observing Figure 13-11, duplicate subsets occur when array elements are selected in different orders, as in the following cases:
|
|
|
|
1. When the first and second rounds select $3$ and $4$ respectively, all subsets containing these two elements are generated, denoted as $[3, 4, \dots]$.
|
|
2. Afterward, when the first round selects $4$, **the second round should skip $3$**, because the subset $[4, 3, \dots]$ generated by this choice is completely duplicate with the subset generated in step `1.`
|
|
|
|
In the search process, each level's choices are tried from left to right, so the rightmost branches are pruned more.
|
|
|
|
1. The first two rounds select $3$ and $5$, generating subset $[3, 5, \dots]$.
|
|
2. The first two rounds select $4$ and $5$, generating subset $[4, 5, \dots]$.
|
|
3. If the first round selects $5$, **the second round should skip $3$ and $4$**, because subsets $[5, 3, \dots]$ and $[5, 4, \dots]$ are completely duplicate with the subsets described in steps `1.` and `2.`
|
|
|
|
{ class="animation-figure" }
|
|
|
|
<p align="center"> Figure 13-11 Different selection orders leading to duplicate subsets </p>
|
|
|
|
In summary, given an input array $[x_1, x_2, \dots, x_n]$, let the selection sequence in the search process be $[x_{i_1}, x_{i_2}, \dots, x_{i_m}]$. This selection sequence must satisfy $i_1 \leq i_2 \leq \dots \leq i_m$; **any selection sequence that does not satisfy this condition will cause duplicates and should be pruned**.
|
|
|
|
### 3. Code Implementation
|
|
|
|
To implement this pruning, we initialize a variable `start` to indicate the starting point of traversal. **After making choice $x_{i}$, set the next round to start traversal from index $i$**. This ensures that the selection sequence satisfies $i_1 \leq i_2 \leq \dots \leq i_m$, guaranteeing subset uniqueness.
|
|
|
|
In addition, we have made the following two optimizations to the code:
|
|
|
|
- Before starting the search, first sort the array `nums`. When traversing all choices, **end the loop immediately when the subset sum exceeds `target`**, because subsequent elements are larger, and their subset sums must exceed `target`.
|
|
- Omit the element sum variable `total` and **use subtraction on `target` to track the sum of elements**. Record the solution when `target` equals $0$.
|
|
|
|
=== "Python"
|
|
|
|
```python title="subset_sum_i.py"
|
|
def backtrack(
|
|
state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
|
|
):
|
|
"""Backtracking algorithm: Subset sum I"""
|
|
# When the subset sum equals target, record the solution
|
|
if target == 0:
|
|
res.append(list(state))
|
|
return
|
|
# Traverse all choices
|
|
# Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for i in range(start, len(choices)):
|
|
# Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
# This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if target - choices[i] < 0:
|
|
break
|
|
# Attempt: make choice, update target, start
|
|
state.append(choices[i])
|
|
# Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i, res)
|
|
# Backtrack: undo choice, restore to previous state
|
|
state.pop()
|
|
|
|
def subset_sum_i(nums: list[int], target: int) -> list[list[int]]:
|
|
"""Solve subset sum I"""
|
|
state = [] # State (subset)
|
|
nums.sort() # Sort nums
|
|
start = 0 # Start point for traversal
|
|
res = [] # Result list (subset list)
|
|
backtrack(state, target, nums, start, res)
|
|
return res
|
|
```
|
|
|
|
=== "C++"
|
|
|
|
```cpp title="subset_sum_i.cpp"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
void backtrack(vector<int> &state, int target, vector<int> &choices, int start, vector<vector<int>> &res) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.push_back(state);
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for (int i = start; i < choices.size(); i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.push_back(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.pop_back();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
vector<vector<int>> subsetSumI(vector<int> &nums, int target) {
|
|
vector<int> state; // State (subset)
|
|
sort(nums.begin(), nums.end()); // Sort nums
|
|
int start = 0; // Start point for traversal
|
|
vector<vector<int>> res; // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "Java"
|
|
|
|
```java title="subset_sum_i.java"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
void backtrack(List<Integer> state, int target, int[] choices, int start, List<List<Integer>> res) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.add(new ArrayList<>(state));
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for (int i = start; i < choices.length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.add(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.remove(state.size() - 1);
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
List<List<Integer>> subsetSumI(int[] nums, int target) {
|
|
List<Integer> state = new ArrayList<>(); // State (subset)
|
|
Arrays.sort(nums); // Sort nums
|
|
int start = 0; // Start point for traversal
|
|
List<List<Integer>> res = new ArrayList<>(); // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "C#"
|
|
|
|
```csharp title="subset_sum_i.cs"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
void Backtrack(List<int> state, int target, int[] choices, int start, List<List<int>> res) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.Add(new List<int>(state));
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for (int i = start; i < choices.Length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.Add(choices[i]);
|
|
// Proceed to the next round of selection
|
|
Backtrack(state, target - choices[i], choices, i, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.RemoveAt(state.Count - 1);
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
List<List<int>> SubsetSumI(int[] nums, int target) {
|
|
List<int> state = []; // State (subset)
|
|
Array.Sort(nums); // Sort nums
|
|
int start = 0; // Start point for traversal
|
|
List<List<int>> res = []; // Result list (subset list)
|
|
Backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "Go"
|
|
|
|
```go title="subset_sum_i.go"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
func backtrackSubsetSumI(start, target int, state, choices *[]int, res *[][]int) {
|
|
// When the subset sum equals target, record the solution
|
|
if target == 0 {
|
|
newState := append([]int{}, *state...)
|
|
*res = append(*res, newState)
|
|
return
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for i := start; i < len(*choices); i++ {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if target-(*choices)[i] < 0 {
|
|
break
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
*state = append(*state, (*choices)[i])
|
|
// Proceed to the next round of selection
|
|
backtrackSubsetSumI(i, target-(*choices)[i], state, choices, res)
|
|
// Backtrack: undo choice, restore to previous state
|
|
*state = (*state)[:len(*state)-1]
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
func subsetSumI(nums []int, target int) [][]int {
|
|
state := make([]int, 0) // State (subset)
|
|
sort.Ints(nums) // Sort nums
|
|
start := 0 // Start point for traversal
|
|
res := make([][]int, 0) // Result list (subset list)
|
|
backtrackSubsetSumI(start, target, &state, &nums, &res)
|
|
return res
|
|
}
|
|
```
|
|
|
|
=== "Swift"
|
|
|
|
```swift title="subset_sum_i.swift"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
func backtrack(state: inout [Int], target: Int, choices: [Int], start: Int, res: inout [[Int]]) {
|
|
// When the subset sum equals target, record the solution
|
|
if target == 0 {
|
|
res.append(state)
|
|
return
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for i in choices.indices.dropFirst(start) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if target - choices[i] < 0 {
|
|
break
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.append(choices[i])
|
|
// Proceed to the next round of selection
|
|
backtrack(state: &state, target: target - choices[i], choices: choices, start: i, res: &res)
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.removeLast()
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
func subsetSumI(nums: [Int], target: Int) -> [[Int]] {
|
|
var state: [Int] = [] // State (subset)
|
|
let nums = nums.sorted() // Sort nums
|
|
let start = 0 // Start point for traversal
|
|
var res: [[Int]] = [] // Result list (subset list)
|
|
backtrack(state: &state, target: target, choices: nums, start: start, res: &res)
|
|
return res
|
|
}
|
|
```
|
|
|
|
=== "JS"
|
|
|
|
```javascript title="subset_sum_i.js"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
function backtrack(state, target, choices, start, res) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target === 0) {
|
|
res.push([...state]);
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for (let i = start; i < choices.length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.push(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.pop();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
function subsetSumI(nums, target) {
|
|
const state = []; // State (subset)
|
|
nums.sort((a, b) => a - b); // Sort nums
|
|
const start = 0; // Start point for traversal
|
|
const res = []; // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "TS"
|
|
|
|
```typescript title="subset_sum_i.ts"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
function backtrack(
|
|
state: number[],
|
|
target: number,
|
|
choices: number[],
|
|
start: number,
|
|
res: number[][]
|
|
): void {
|
|
// When the subset sum equals target, record the solution
|
|
if (target === 0) {
|
|
res.push([...state]);
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for (let i = start; i < choices.length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.push(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.pop();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
function subsetSumI(nums: number[], target: number): number[][] {
|
|
const state = []; // State (subset)
|
|
nums.sort((a, b) => a - b); // Sort nums
|
|
const start = 0; // Start point for traversal
|
|
const res = []; // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "Dart"
|
|
|
|
```dart title="subset_sum_i.dart"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
void backtrack(
|
|
List<int> state,
|
|
int target,
|
|
List<int> choices,
|
|
int start,
|
|
List<List<int>> res,
|
|
) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.add(List.from(state));
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for (int i = start; i < choices.length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.add(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.removeLast();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
List<List<int>> subsetSumI(List<int> nums, int target) {
|
|
List<int> state = []; // State (subset)
|
|
nums.sort(); // Sort nums
|
|
int start = 0; // Start point for traversal
|
|
List<List<int>> res = []; // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "Rust"
|
|
|
|
```rust title="subset_sum_i.rs"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
fn backtrack(
|
|
state: &mut Vec<i32>,
|
|
target: i32,
|
|
choices: &[i32],
|
|
start: usize,
|
|
res: &mut Vec<Vec<i32>>,
|
|
) {
|
|
// When the subset sum equals target, record the solution
|
|
if target == 0 {
|
|
res.push(state.clone());
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for i in start..choices.len() {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if target - choices[i] < 0 {
|
|
break;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.push(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.pop();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
fn subset_sum_i(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
|
|
let mut state = Vec::new(); // State (subset)
|
|
nums.sort(); // Sort nums
|
|
let start = 0; // Start point for traversal
|
|
let mut res = Vec::new(); // Result list (subset list)
|
|
backtrack(&mut state, target, nums, start, &mut res);
|
|
res
|
|
}
|
|
```
|
|
|
|
=== "C"
|
|
|
|
```c title="subset_sum_i.c"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
void backtrack(int target, int *choices, int choicesSize, int start) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
for (int i = 0; i < stateSize; ++i) {
|
|
res[resSize][i] = state[i];
|
|
}
|
|
resColSizes[resSize++] = stateSize;
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for (int i = start; i < choicesSize; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state[stateSize] = choices[i];
|
|
stateSize++;
|
|
// Proceed to the next round of selection
|
|
backtrack(target - choices[i], choices, choicesSize, i);
|
|
// Backtrack: undo choice, restore to previous state
|
|
stateSize--;
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
void subsetSumI(int *nums, int numsSize, int target) {
|
|
qsort(nums, numsSize, sizeof(int), cmp); // Sort nums
|
|
int start = 0; // Start point for traversal
|
|
backtrack(target, nums, numsSize, start);
|
|
}
|
|
```
|
|
|
|
=== "Kotlin"
|
|
|
|
```kotlin title="subset_sum_i.kt"
|
|
/* Backtracking algorithm: Subset sum I */
|
|
fun backtrack(
|
|
state: MutableList<Int>,
|
|
target: Int,
|
|
choices: IntArray,
|
|
start: Int,
|
|
res: MutableList<MutableList<Int>?>
|
|
) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.add(state.toMutableList())
|
|
return
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for (i in start..<choices.size) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.add(choices[i])
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i, res)
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.removeAt(state.size - 1)
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum I */
|
|
fun subsetSumI(nums: IntArray, target: Int): MutableList<MutableList<Int>?> {
|
|
val state = mutableListOf<Int>() // State (subset)
|
|
nums.sort() // Sort nums
|
|
val start = 0 // Start point for traversal
|
|
val res = mutableListOf<MutableList<Int>?>() // Result list (subset list)
|
|
backtrack(state, target, nums, start, res)
|
|
return res
|
|
}
|
|
```
|
|
|
|
=== "Ruby"
|
|
|
|
```ruby title="subset_sum_i.rb"
|
|
### Backtracking: subset sum I ###
|
|
def backtrack(state, target, choices, start, res)
|
|
# When the subset sum equals target, record the solution
|
|
if target.zero?
|
|
res << state.dup
|
|
return
|
|
end
|
|
# Traverse all choices
|
|
# Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
for i in start...choices.length
|
|
# Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
# This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
break if target - choices[i] < 0
|
|
# Attempt: make choice, update target, start
|
|
state << choices[i]
|
|
# Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i, res)
|
|
# Backtrack: undo choice, restore to previous state
|
|
state.pop
|
|
end
|
|
end
|
|
|
|
### Solve subset sum I ###
|
|
def subset_sum_i(nums, target)
|
|
state = [] # State (subset)
|
|
nums.sort! # Sort nums
|
|
start = 0 # Start point for traversal
|
|
res = [] # Result list (subset list)
|
|
backtrack(state, target, nums, start, res)
|
|
res
|
|
end
|
|
```
|
|
|
|
Figure 13-12 shows the complete backtracking process when array $[3, 4, 5]$ and target element $9$ are input to the above code.
|
|
|
|
{ class="animation-figure" }
|
|
|
|
<p align="center"> Figure 13-12 Subset-sum I backtracking process </p>
|
|
|
|
## 13.3.2 With Duplicate Elements in Array
|
|
|
|
!!! question
|
|
|
|
Given a positive integer array `nums` and a target positive integer `target`, find all possible combinations where the sum of elements in the combination equals `target`. **The given array may contain duplicate elements, and each element can be selected at most once**. Return these combinations in list form, where the list should not contain duplicate combinations.
|
|
|
|
Compared to the previous problem, **the input array in this problem may contain duplicate elements**, which introduces new challenges. For example, given array $[4, \hat{4}, 5]$ and target element $9$, the output of the existing code is $[4, 5], [\hat{4}, 5]$, which contains duplicate subsets.
|
|
|
|
**The reason for this duplication is that equal elements are selected multiple times in a certain round**. In Figure 13-13, the first round has three choices, two of which are $4$, creating two duplicate search branches that output duplicate subsets. Similarly, the two $4$'s in the second round also produce duplicate subsets.
|
|
|
|
{ class="animation-figure" }
|
|
|
|
<p align="center"> Figure 13-13 Duplicate subsets caused by equal elements </p>
|
|
|
|
### 1. Pruning Equal Elements
|
|
|
|
To solve this problem, **we need to limit equal elements to be selected only once in each round**. The implementation is quite clever: since the array is already sorted, equal elements are adjacent. This means that in a certain round of selection, if the current element equals the element to its left, it means this element has already been selected, so we skip the current element directly.
|
|
|
|
At the same time, **this problem specifies that each array element can only be selected once**. Fortunately, we can also use the variable `start` to satisfy this constraint: after making choice $x_{i}$, set the next round to start traversal from index $i + 1$ onwards. This both eliminates duplicate subsets and avoids selecting elements multiple times.
|
|
|
|
### 2. Code Implementation
|
|
|
|
=== "Python"
|
|
|
|
```python title="subset_sum_ii.py"
|
|
def backtrack(
|
|
state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
|
|
):
|
|
"""Backtracking algorithm: Subset sum II"""
|
|
# When the subset sum equals target, record the solution
|
|
if target == 0:
|
|
res.append(list(state))
|
|
return
|
|
# Traverse all choices
|
|
# Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
# Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for i in range(start, len(choices)):
|
|
# Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
# This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if target - choices[i] < 0:
|
|
break
|
|
# Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if i > start and choices[i] == choices[i - 1]:
|
|
continue
|
|
# Attempt: make choice, update target, start
|
|
state.append(choices[i])
|
|
# Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i + 1, res)
|
|
# Backtrack: undo choice, restore to previous state
|
|
state.pop()
|
|
|
|
def subset_sum_ii(nums: list[int], target: int) -> list[list[int]]:
|
|
"""Solve subset sum II"""
|
|
state = [] # State (subset)
|
|
nums.sort() # Sort nums
|
|
start = 0 # Start point for traversal
|
|
res = [] # Result list (subset list)
|
|
backtrack(state, target, nums, start, res)
|
|
return res
|
|
```
|
|
|
|
=== "C++"
|
|
|
|
```cpp title="subset_sum_ii.cpp"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
void backtrack(vector<int> &state, int target, vector<int> &choices, int start, vector<vector<int>> &res) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.push_back(state);
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for (int i = start; i < choices.size(); i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if (i > start && choices[i] == choices[i - 1]) {
|
|
continue;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.push_back(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i + 1, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.pop_back();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
vector<vector<int>> subsetSumII(vector<int> &nums, int target) {
|
|
vector<int> state; // State (subset)
|
|
sort(nums.begin(), nums.end()); // Sort nums
|
|
int start = 0; // Start point for traversal
|
|
vector<vector<int>> res; // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "Java"
|
|
|
|
```java title="subset_sum_ii.java"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
void backtrack(List<Integer> state, int target, int[] choices, int start, List<List<Integer>> res) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.add(new ArrayList<>(state));
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for (int i = start; i < choices.length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if (i > start && choices[i] == choices[i - 1]) {
|
|
continue;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.add(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i + 1, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.remove(state.size() - 1);
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
List<List<Integer>> subsetSumII(int[] nums, int target) {
|
|
List<Integer> state = new ArrayList<>(); // State (subset)
|
|
Arrays.sort(nums); // Sort nums
|
|
int start = 0; // Start point for traversal
|
|
List<List<Integer>> res = new ArrayList<>(); // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "C#"
|
|
|
|
```csharp title="subset_sum_ii.cs"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
void Backtrack(List<int> state, int target, int[] choices, int start, List<List<int>> res) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.Add(new List<int>(state));
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for (int i = start; i < choices.Length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if (i > start && choices[i] == choices[i - 1]) {
|
|
continue;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.Add(choices[i]);
|
|
// Proceed to the next round of selection
|
|
Backtrack(state, target - choices[i], choices, i + 1, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.RemoveAt(state.Count - 1);
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
List<List<int>> SubsetSumII(int[] nums, int target) {
|
|
List<int> state = []; // State (subset)
|
|
Array.Sort(nums); // Sort nums
|
|
int start = 0; // Start point for traversal
|
|
List<List<int>> res = []; // Result list (subset list)
|
|
Backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "Go"
|
|
|
|
```go title="subset_sum_ii.go"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
func backtrackSubsetSumII(start, target int, state, choices *[]int, res *[][]int) {
|
|
// When the subset sum equals target, record the solution
|
|
if target == 0 {
|
|
newState := append([]int{}, *state...)
|
|
*res = append(*res, newState)
|
|
return
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for i := start; i < len(*choices); i++ {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if target-(*choices)[i] < 0 {
|
|
break
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if i > start && (*choices)[i] == (*choices)[i-1] {
|
|
continue
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
*state = append(*state, (*choices)[i])
|
|
// Proceed to the next round of selection
|
|
backtrackSubsetSumII(i+1, target-(*choices)[i], state, choices, res)
|
|
// Backtrack: undo choice, restore to previous state
|
|
*state = (*state)[:len(*state)-1]
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
func subsetSumII(nums []int, target int) [][]int {
|
|
state := make([]int, 0) // State (subset)
|
|
sort.Ints(nums) // Sort nums
|
|
start := 0 // Start point for traversal
|
|
res := make([][]int, 0) // Result list (subset list)
|
|
backtrackSubsetSumII(start, target, &state, &nums, &res)
|
|
return res
|
|
}
|
|
```
|
|
|
|
=== "Swift"
|
|
|
|
```swift title="subset_sum_ii.swift"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
func backtrack(state: inout [Int], target: Int, choices: [Int], start: Int, res: inout [[Int]]) {
|
|
// When the subset sum equals target, record the solution
|
|
if target == 0 {
|
|
res.append(state)
|
|
return
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for i in choices.indices.dropFirst(start) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if target - choices[i] < 0 {
|
|
break
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if i > start, choices[i] == choices[i - 1] {
|
|
continue
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.append(choices[i])
|
|
// Proceed to the next round of selection
|
|
backtrack(state: &state, target: target - choices[i], choices: choices, start: i + 1, res: &res)
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.removeLast()
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
func subsetSumII(nums: [Int], target: Int) -> [[Int]] {
|
|
var state: [Int] = [] // State (subset)
|
|
let nums = nums.sorted() // Sort nums
|
|
let start = 0 // Start point for traversal
|
|
var res: [[Int]] = [] // Result list (subset list)
|
|
backtrack(state: &state, target: target, choices: nums, start: start, res: &res)
|
|
return res
|
|
}
|
|
```
|
|
|
|
=== "JS"
|
|
|
|
```javascript title="subset_sum_ii.js"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
function backtrack(state, target, choices, start, res) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target === 0) {
|
|
res.push([...state]);
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for (let i = start; i < choices.length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if (i > start && choices[i] === choices[i - 1]) {
|
|
continue;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.push(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i + 1, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.pop();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
function subsetSumII(nums, target) {
|
|
const state = []; // State (subset)
|
|
nums.sort((a, b) => a - b); // Sort nums
|
|
const start = 0; // Start point for traversal
|
|
const res = []; // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "TS"
|
|
|
|
```typescript title="subset_sum_ii.ts"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
function backtrack(
|
|
state: number[],
|
|
target: number,
|
|
choices: number[],
|
|
start: number,
|
|
res: number[][]
|
|
): void {
|
|
// When the subset sum equals target, record the solution
|
|
if (target === 0) {
|
|
res.push([...state]);
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for (let i = start; i < choices.length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if (i > start && choices[i] === choices[i - 1]) {
|
|
continue;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.push(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i + 1, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.pop();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
function subsetSumII(nums: number[], target: number): number[][] {
|
|
const state = []; // State (subset)
|
|
nums.sort((a, b) => a - b); // Sort nums
|
|
const start = 0; // Start point for traversal
|
|
const res = []; // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "Dart"
|
|
|
|
```dart title="subset_sum_ii.dart"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
void backtrack(
|
|
List<int> state,
|
|
int target,
|
|
List<int> choices,
|
|
int start,
|
|
List<List<int>> res,
|
|
) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.add(List.from(state));
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for (int i = start; i < choices.length; i++) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break;
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if (i > start && choices[i] == choices[i - 1]) {
|
|
continue;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.add(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i + 1, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.removeLast();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
List<List<int>> subsetSumII(List<int> nums, int target) {
|
|
List<int> state = []; // State (subset)
|
|
nums.sort(); // Sort nums
|
|
int start = 0; // Start point for traversal
|
|
List<List<int>> res = []; // Result list (subset list)
|
|
backtrack(state, target, nums, start, res);
|
|
return res;
|
|
}
|
|
```
|
|
|
|
=== "Rust"
|
|
|
|
```rust title="subset_sum_ii.rs"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
fn backtrack(
|
|
state: &mut Vec<i32>,
|
|
target: i32,
|
|
choices: &[i32],
|
|
start: usize,
|
|
res: &mut Vec<Vec<i32>>,
|
|
) {
|
|
// When the subset sum equals target, record the solution
|
|
if target == 0 {
|
|
res.push(state.clone());
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for i in start..choices.len() {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if target - choices[i] < 0 {
|
|
break;
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if i > start && choices[i] == choices[i - 1] {
|
|
continue;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.push(choices[i]);
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i + 1, res);
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.pop();
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
fn subset_sum_ii(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
|
|
let mut state = Vec::new(); // State (subset)
|
|
nums.sort(); // Sort nums
|
|
let start = 0; // Start point for traversal
|
|
let mut res = Vec::new(); // Result list (subset list)
|
|
backtrack(&mut state, target, nums, start, &mut res);
|
|
res
|
|
}
|
|
```
|
|
|
|
=== "C"
|
|
|
|
```c title="subset_sum_ii.c"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
void backtrack(int target, int *choices, int choicesSize, int start) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
for (int i = 0; i < stateSize; i++) {
|
|
res[resSize][i] = state[i];
|
|
}
|
|
resColSizes[resSize++] = stateSize;
|
|
return;
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for (int i = start; i < choicesSize; i++) {
|
|
// Pruning 1: Skip if subset sum exceeds target
|
|
if (target - choices[i] < 0) {
|
|
continue;
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if (i > start && choices[i] == choices[i - 1]) {
|
|
continue;
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state[stateSize] = choices[i];
|
|
stateSize++;
|
|
// Proceed to the next round of selection
|
|
backtrack(target - choices[i], choices, choicesSize, i + 1);
|
|
// Backtrack: undo choice, restore to previous state
|
|
stateSize--;
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
void subsetSumII(int *nums, int numsSize, int target) {
|
|
// Sort nums
|
|
qsort(nums, numsSize, sizeof(int), cmp);
|
|
// Start backtracking
|
|
backtrack(target, nums, numsSize, 0);
|
|
}
|
|
```
|
|
|
|
=== "Kotlin"
|
|
|
|
```kotlin title="subset_sum_ii.kt"
|
|
/* Backtracking algorithm: Subset sum II */
|
|
fun backtrack(
|
|
state: MutableList<Int>,
|
|
target: Int,
|
|
choices: IntArray,
|
|
start: Int,
|
|
res: MutableList<MutableList<Int>?>
|
|
) {
|
|
// When the subset sum equals target, record the solution
|
|
if (target == 0) {
|
|
res.add(state.toMutableList())
|
|
return
|
|
}
|
|
// Traverse all choices
|
|
// Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
// Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for (i in start..<choices.size) {
|
|
// Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
if (target - choices[i] < 0) {
|
|
break
|
|
}
|
|
// Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
if (i > start && choices[i] == choices[i - 1]) {
|
|
continue
|
|
}
|
|
// Attempt: make choice, update target, start
|
|
state.add(choices[i])
|
|
// Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i + 1, res)
|
|
// Backtrack: undo choice, restore to previous state
|
|
state.removeAt(state.size - 1)
|
|
}
|
|
}
|
|
|
|
/* Solve subset sum II */
|
|
fun subsetSumII(nums: IntArray, target: Int): MutableList<MutableList<Int>?> {
|
|
val state = mutableListOf<Int>() // State (subset)
|
|
nums.sort() // Sort nums
|
|
val start = 0 // Start point for traversal
|
|
val res = mutableListOf<MutableList<Int>?>() // Result list (subset list)
|
|
backtrack(state, target, nums, start, res)
|
|
return res
|
|
}
|
|
```
|
|
|
|
=== "Ruby"
|
|
|
|
```ruby title="subset_sum_ii.rb"
|
|
### Backtracking: subset sum II ###
|
|
def backtrack(state, target, choices, start, res)
|
|
# When the subset sum equals target, record the solution
|
|
if target.zero?
|
|
res << state.dup
|
|
return
|
|
end
|
|
|
|
# Traverse all choices
|
|
# Pruning 2: start traversing from start to avoid generating duplicate subsets
|
|
# Pruning 3: start traversing from start to avoid repeatedly selecting the same element
|
|
for i in start...choices.length
|
|
# Pruning 1: if the subset sum exceeds target, end the loop directly
|
|
# This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
|
|
break if target - choices[i] < 0
|
|
# Pruning 4: if this element equals the left element, it means this search branch is duplicate, skip it directly
|
|
next if i > start && choices[i] == choices[i - 1]
|
|
# Attempt: make choice, update target, start
|
|
state << choices[i]
|
|
# Proceed to the next round of selection
|
|
backtrack(state, target - choices[i], choices, i + 1, res)
|
|
# Backtrack: undo choice, restore to previous state
|
|
state.pop
|
|
end
|
|
end
|
|
|
|
### Solve subset sum II ###
|
|
def subset_sum_ii(nums, target)
|
|
state = [] # State (subset)
|
|
nums.sort! # Sort nums
|
|
start = 0 # Start point for traversal
|
|
res = [] # Result list (subset list)
|
|
backtrack(state, target, nums, start, res)
|
|
res
|
|
end
|
|
```
|
|
|
|
Figure 13-14 shows the backtracking process for array $[4, 4, 5]$ and target element $9$, which includes four types of pruning operations. Combine the illustration with the code comments to understand the entire search process and how each pruning operation works.
|
|
|
|
{ class="animation-figure" }
|
|
|
|
<p align="center"> Figure 13-14 Subset-sum II backtracking process </p>
|