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---
comments: true
---
# 14.2   Characteristics of dynamic programming problems
In the previous section, we learned how dynamic programming solves the original problem by decomposing it into subproblems. In fact, subproblem decomposition is a general algorithmic approach, with different emphases in divide and conquer, dynamic programming, and backtracking.
- Divide and conquer algorithms recursively divide the original problem into multiple independent subproblems until the smallest subproblems are reached, and combine the solutions of the subproblems during backtracking to ultimately obtain the solution to the original problem.
- Dynamic programming also decomposes the problem recursively, but the main difference from divide and conquer algorithms is that the subproblems in dynamic programming are interdependent, and many overlapping subproblems will appear during the decomposition process.
- Backtracking algorithms exhaust all possible solutions through trial and error and avoid unnecessary search branches by pruning. The solution to the original problem consists of a series of decision steps, and we can consider each sub-sequence before each decision step as a subproblem.
In fact, dynamic programming is commonly used to solve optimization problems, which not only include overlapping subproblems but also have two other major characteristics: optimal substructure and statelessness.
## 14.2.1   Optimal substructure
We make a slight modification to the stair climbing problem to make it more suitable to demonstrate the concept of optimal substructure.
!!! question "Minimum cost of climbing stairs"
Given a staircase, you can step up 1 or 2 steps at a time, and each step on the staircase has a non-negative integer representing the cost you need to pay at that step. Given a non-negative integer array $cost$, where $cost[i]$ represents the cost you need to pay at the $i$-th step, $cost[0]$ is the ground (starting point). What is the minimum cost required to reach the top?
As shown in the Figure 14-6 , if the costs of the 1st, 2nd, and 3rd steps are $1$, $10$, and $1$ respectively, then the minimum cost to climb to the 3rd step from the ground is $2$.
![Minimum cost to climb to the 3rd step](dp_problem_features.assets/min_cost_cs_example.png){ class="animation-figure" }
<p align="center"> Figure 14-6 &nbsp; Minimum cost to climb to the 3rd step </p>
Let $dp[i]$ be the cumulative cost of climbing to the $i$-th step. Since the $i$-th step can only come from the $i-1$ or $i-2$ step, $dp[i]$ can only be either $dp[i-1] + cost[i]$ or $dp[i-2] + cost[i]$. To minimize the cost, we should choose the smaller of the two:
$$
dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
$$
This leads us to the meaning of optimal substructure: **The optimal solution to the original problem is constructed from the optimal solutions of subproblems**.
This problem obviously has optimal substructure: we select the better one from the optimal solutions of the two subproblems, $dp[i-1]$ and $dp[i-2]$, and use it to construct the optimal solution for the original problem $dp[i]$.
So, does the stair climbing problem from the previous section have optimal substructure? Its goal is to solve for the number of solutions, which seems to be a counting problem, but if we ask in another way: "Solve for the maximum number of solutions". We surprisingly find that **although the problem has changed, the optimal substructure has emerged**: the maximum number of solutions at the $n$-th step equals the sum of the maximum number of solutions at the $n-1$ and $n-2$ steps. Thus, the interpretation of optimal substructure is quite flexible and will have different meanings in different problems.
According to the state transition equation, and the initial states $dp[1] = cost[1]$ and $dp[2] = cost[2]$, we can obtain the dynamic programming code:
=== "Python"
```python title="min_cost_climbing_stairs_dp.py"
def min_cost_climbing_stairs_dp(cost: list[int]) -> int:
"""爬楼梯最小代价：动态规划"""
n = len(cost) - 1
if n == 1 or n == 2:
return cost[n]
# 初始化 dp 表，用于存储子问题的解
dp = [0] * (n + 1)
# 初始状态：预设最小子问题的解
dp[1], dp[2] = cost[1], cost[2]
# 状态转移：从较小子问题逐步求解较大子问题
for i in range(3, n + 1):
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
return dp[n]
```
=== "C++"
```cpp title="min_cost_climbing_stairs_dp.cpp"
/* 爬楼梯最小代价：动态规划 */
int minCostClimbingStairsDP(vector<int> &cost) {
int n = cost.size() - 1;
if (n == 1 || n == 2)
return cost[n];
// 初始化 dp 表，用于存储子问题的解
vector<int> dp(n + 1);
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "Java"
```java title="min_cost_climbing_stairs_dp.java"
/* 爬楼梯最小代价：动态规划 */
int minCostClimbingStairsDP(int[] cost) {
int n = cost.length - 1;
if (n == 1 || n == 2)
return cost[n];
// 初始化 dp 表，用于存储子问题的解
int[] dp = new int[n + 1];
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "C#"
```csharp title="min_cost_climbing_stairs_dp.cs"
/* 爬楼梯最小代价：动态规划 */
int MinCostClimbingStairsDP(int[] cost) {
int n = cost.Length - 1;
if (n == 1 || n == 2)
return cost[n];
// 初始化 dp 表，用于存储子问题的解
int[] dp = new int[n + 1];
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i] = Math.Min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "Go"
```go title="min_cost_climbing_stairs_dp.go"
/* 爬楼梯最小代价：动态规划 */
func minCostClimbingStairsDP(cost []int) int {
n := len(cost) - 1
if n == 1 || n == 2 {
return cost[n]
}
min := func(a, b int) int {
if a < b {
return a
}
return b
}
// 初始化 dp 表，用于存储子问题的解
dp := make([]int, n+1)
// 初始状态：预设最小子问题的解
dp[1] = cost[1]
dp[2] = cost[2]
// 状态转移：从较小子问题逐步求解较大子问题
for i := 3; i <= n; i++ {
dp[i] = min(dp[i-1], dp[i-2]) + cost[i]
}
return dp[n]
}
```
=== "Swift"
```swift title="min_cost_climbing_stairs_dp.swift"
/* 爬楼梯最小代价：动态规划 */
func minCostClimbingStairsDP(cost: [Int]) -> Int {
let n = cost.count - 1
if n == 1 || n == 2 {
return cost[n]
}
// 初始化 dp 表，用于存储子问题的解
var dp = Array(repeating: 0, count: n + 1)
// 初始状态：预设最小子问题的解
dp[1] = cost[1]
dp[2] = cost[2]
// 状态转移：从较小子问题逐步求解较大子问题
for i in 3 ... n {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
}
return dp[n]
}
```
=== "JS"
```javascript title="min_cost_climbing_stairs_dp.js"
/* 爬楼梯最小代价：动态规划 */
function minCostClimbingStairsDP(cost) {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
// 初始化 dp 表，用于存储子问题的解
const dp = new Array(n + 1);
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (let i = 3; i <= n; i++) {
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "TS"
```typescript title="min_cost_climbing_stairs_dp.ts"
/* 爬楼梯最小代价：动态规划 */
function minCostClimbingStairsDP(cost: Array<number>): number {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
// 初始化 dp 表，用于存储子问题的解
const dp = new Array(n + 1);
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (let i = 3; i <= n; i++) {
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "Dart"
```dart title="min_cost_climbing_stairs_dp.dart"
/* 爬楼梯最小代价：动态规划 */
int minCostClimbingStairsDP(List<int> cost) {
int n = cost.length - 1;
if (n == 1 || n == 2) return cost[n];
// 初始化 dp 表，用于存储子问题的解
List<int> dp = List.filled(n + 1, 0);
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
=== "Rust"
```rust title="min_cost_climbing_stairs_dp.rs"
/* 爬楼梯最小代价：动态规划 */
fn min_cost_climbing_stairs_dp(cost: &[i32]) -> i32 {
let n = cost.len() - 1;
if n == 1 || n == 2 {
return cost[n];
}
// 初始化 dp 表，用于存储子问题的解
let mut dp = vec![-1; n + 1];
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for i in 3..=n {
dp[i] = cmp::min(dp[i - 1], dp[i - 2]) + cost[i];
}
dp[n]
}
```
=== "C"
```c title="min_cost_climbing_stairs_dp.c"
/* 爬楼梯最小代价：动态规划 */
int minCostClimbingStairsDP(int cost[], int costSize) {
int n = costSize - 1;
if (n == 1 || n == 2)
return cost[n];
// 初始化 dp 表，用于存储子问题的解
int *dp = calloc(n + 1, sizeof(int));
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i] = myMin(dp[i - 1], dp[i - 2]) + cost[i];
}
int res = dp[n];
// 释放内存
free(dp);
return res;
}
```
=== "Kotlin"
```kotlin title="min_cost_climbing_stairs_dp.kt"
/* 爬楼梯最小代价：动态规划 */
fun minCostClimbingStairsDP(cost: IntArray): Int {
val n = cost.size - 1
if (n == 1 || n == 2) return cost[n]
// 初始化 dp 表，用于存储子问题的解
val dp = IntArray(n + 1)
// 初始状态：预设最小子问题的解
dp[1] = cost[1]
dp[2] = cost[2]
// 状态转移：从较小子问题逐步求解较大子问题
for (i in 3..n) {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
}
return dp[n]
}
```
=== "Ruby"
```ruby title="min_cost_climbing_stairs_dp.rb"
[class]{}-[func]{min_cost_climbing_stairs_dp}
```
=== "Zig"
```zig title="min_cost_climbing_stairs_dp.zig"
// 爬楼梯最小代价：动态规划
fn minCostClimbingStairsDP(comptime cost: []i32) i32 {
comptime var n = cost.len - 1;
if (n == 1 or n == 2) {
return cost[n];
}
// 初始化 dp 表，用于存储子问题的解
var dp = [_]i32{-1} ** (n + 1);
// 初始状态：预设最小子问题的解
dp[1] = cost[1];
dp[2] = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (3..n + 1) |i| {
dp[i] = @min(dp[i - 1], dp[i - 2]) + cost[i];
}
return dp[n];
}
```
??? pythontutor "Code Visualization"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20min%28dp%5Bi%20-%201%5D,%20dp%5Bi%20-%202%5D%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20min%28dp%5Bi%20-%201%5D,%20dp%5Bi%20-%202%5D%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
The Figure 14-7 shows the dynamic programming process for the above code.
![Dynamic programming process for minimum cost of climbing stairs](dp_problem_features.assets/min_cost_cs_dp.png){ class="animation-figure" }
<p align="center"> Figure 14-7 &nbsp; Dynamic programming process for minimum cost of climbing stairs </p>
This problem can also be space-optimized, compressing one dimension to zero, reducing the space complexity from $O(n)$ to $O(1)$:
=== "Python"
```python title="min_cost_climbing_stairs_dp.py"
def min_cost_climbing_stairs_dp_comp(cost: list[int]) -> int:
"""爬楼梯最小代价：空间优化后的动态规划"""
n = len(cost) - 1
if n == 1 or n == 2:
return cost[n]
a, b = cost[1], cost[2]
for i in range(3, n + 1):
a, b = b, min(a, b) + cost[i]
return b
```
=== "C++"
```cpp title="min_cost_climbing_stairs_dp.cpp"
/* 爬楼梯最小代价：空间优化后的动态规划 */
int minCostClimbingStairsDPComp(vector<int> &cost) {
int n = cost.size() - 1;
if (n == 1 || n == 2)
return cost[n];
int a = cost[1], b = cost[2];
for (int i = 3; i <= n; i++) {
int tmp = b;
b = min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "Java"
```java title="min_cost_climbing_stairs_dp.java"
/* 爬楼梯最小代价：空间优化后的动态规划 */
int minCostClimbingStairsDPComp(int[] cost) {
int n = cost.length - 1;
if (n == 1 || n == 2)
return cost[n];
int a = cost[1], b = cost[2];
for (int i = 3; i <= n; i++) {
int tmp = b;
b = Math.min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "C#"
```csharp title="min_cost_climbing_stairs_dp.cs"
/* 爬楼梯最小代价：空间优化后的动态规划 */
int MinCostClimbingStairsDPComp(int[] cost) {
int n = cost.Length - 1;
if (n == 1 || n == 2)
return cost[n];
int a = cost[1], b = cost[2];
for (int i = 3; i <= n; i++) {
int tmp = b;
b = Math.Min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "Go"
```go title="min_cost_climbing_stairs_dp.go"
/* 爬楼梯最小代价：空间优化后的动态规划 */
func minCostClimbingStairsDPComp(cost []int) int {
n := len(cost) - 1
if n == 1 || n == 2 {
return cost[n]
}
min := func(a, b int) int {
if a < b {
return a
}
return b
}
// 初始状态：预设最小子问题的解
a, b := cost[1], cost[2]
// 状态转移：从较小子问题逐步求解较大子问题
for i := 3; i <= n; i++ {
tmp := b
b = min(a, tmp) + cost[i]
a = tmp
}
return b
}
```
=== "Swift"
```swift title="min_cost_climbing_stairs_dp.swift"
/* 爬楼梯最小代价：空间优化后的动态规划 */
func minCostClimbingStairsDPComp(cost: [Int]) -> Int {
let n = cost.count - 1
if n == 1 || n == 2 {
return cost[n]
}
var (a, b) = (cost[1], cost[2])
for i in 3 ... n {
(a, b) = (b, min(a, b) + cost[i])
}
return b
}
```
=== "JS"
```javascript title="min_cost_climbing_stairs_dp.js"
/* 爬楼梯最小代价：状态压缩后的动态规划 */
function minCostClimbingStairsDPComp(cost) {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
let a = cost[1],
b = cost[2];
for (let i = 3; i <= n; i++) {
const tmp = b;
b = Math.min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "TS"
```typescript title="min_cost_climbing_stairs_dp.ts"
/* 爬楼梯最小代价：状态压缩后的动态规划 */
function minCostClimbingStairsDPComp(cost: Array<number>): number {
const n = cost.length - 1;
if (n === 1 || n === 2) {
return cost[n];
}
let a = cost[1],
b = cost[2];
for (let i = 3; i <= n; i++) {
const tmp = b;
b = Math.min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "Dart"
```dart title="min_cost_climbing_stairs_dp.dart"
/* 爬楼梯最小代价：空间优化后的动态规划 */
int minCostClimbingStairsDPComp(List<int> cost) {
int n = cost.length - 1;
if (n == 1 || n == 2) return cost[n];
int a = cost[1], b = cost[2];
for (int i = 3; i <= n; i++) {
int tmp = b;
b = min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "Rust"
```rust title="min_cost_climbing_stairs_dp.rs"
/* 爬楼梯最小代价：空间优化后的动态规划 */
fn min_cost_climbing_stairs_dp_comp(cost: &[i32]) -> i32 {
let n = cost.len() - 1;
if n == 1 || n == 2 {
return cost[n];
};
let (mut a, mut b) = (cost[1], cost[2]);
for i in 3..=n {
let tmp = b;
b = cmp::min(a, tmp) + cost[i];
a = tmp;
}
b
}
```
=== "C"
```c title="min_cost_climbing_stairs_dp.c"
/* 爬楼梯最小代价：空间优化后的动态规划 */
int minCostClimbingStairsDPComp(int cost[], int costSize) {
int n = costSize - 1;
if (n == 1 || n == 2)
return cost[n];
int a = cost[1], b = cost[2];
for (int i = 3; i <= n; i++) {
int tmp = b;
b = myMin(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
=== "Kotlin"
```kotlin title="min_cost_climbing_stairs_dp.kt"
/* 爬楼梯最小代价：空间优化后的动态规划 */
fun minCostClimbingStairsDPComp(cost: IntArray): Int {
val n = cost.size - 1
if (n == 1 || n == 2) return cost[n]
var a = cost[1]
var b = cost[2]
for (i in 3..n) {
val tmp = b
b = min(a, tmp) + cost[i]
a = tmp
}
return b
}
```
=== "Ruby"
```ruby title="min_cost_climbing_stairs_dp.rb"
[class]{}-[func]{min_cost_climbing_stairs_dp_comp}
```
=== "Zig"
```zig title="min_cost_climbing_stairs_dp.zig"
// 爬楼梯最小代价：空间优化后的动态规划
fn minCostClimbingStairsDPComp(cost: []i32) i32 {
var n = cost.len - 1;
if (n == 1 or n == 2) {
return cost[n];
}
var a = cost[1];
var b = cost[2];
// 状态转移：从较小子问题逐步求解较大子问题
for (3..n + 1) |i| {
var tmp = b;
b = @min(a, tmp) + cost[i];
a = tmp;
}
return b;
}
```
??? pythontutor "Code Visualization"
<div style="height: 513px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp_comp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20a,%20b%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a,%20b%20%3D%20b,%20min%28a,%20b%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20b%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp_comp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp_comp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20a,%20b%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a,%20b%20%3D%20b,%20min%28a,%20b%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20b%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp_comp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
## 14.2.2 &nbsp; Statelessness
Statelessness is one of the important characteristics that make dynamic programming effective in solving problems. Its definition is: **Given a certain state, its future development is only related to the current state and unrelated to all past states experienced**.
Taking the stair climbing problem as an example, given state $i$, it will develop into states $i+1$ and $i+2$, corresponding to jumping 1 step and 2 steps respectively. When making these two choices, we do not need to consider the states before state $i$, as they do not affect the future of state $i$.
However, if we add a constraint to the stair climbing problem, the situation changes.
!!! question "Stair climbing with constraints"
Given a staircase with $n$ steps, you can go up 1 or 2 steps each time, **but you cannot jump 1 step twice in a row**. How many ways are there to climb to the top?
As shown in the Figure 14-8 , there are only 2 feasible options for climbing to the 3rd step, among which the option of jumping 1 step three times in a row does not meet the constraint condition and is therefore discarded.
![Number of feasible options for climbing to the 3rd step with constraints](dp_problem_features.assets/climbing_stairs_constraint_example.png){ class="animation-figure" }
<p align="center"> Figure 14-8 &nbsp; Number of feasible options for climbing to the 3rd step with constraints </p>
In this problem, if the last round was a jump of 1 step, then the next round must be a jump of 2 steps. This means that **the next step choice cannot be independently determined by the current state (current stair step), but also depends on the previous state (last round's stair step)**.
It is not difficult to find that this problem no longer satisfies statelessness, and the state transition equation $dp[i] = dp[i-1] + dp[i-2]$ also fails, because $dp[i-1]$ represents this round's jump of 1 step, but it includes many "last round was a jump of 1 step" options, which, to meet the constraint, cannot be directly included in $dp[i]$.
For this, we need to expand the state definition: **State $[i, j]$ represents being on the $i$-th step and the last round was a jump of $j$ steps**, where $j \in \{1, 2\}$. This state definition effectively distinguishes whether the last round was a jump of 1 step or 2 steps, and we can judge accordingly where the current state came from.
- When the last round was a jump of 1 step, the round before last could only choose to jump 2 steps, that is, $dp[i, 1]$ can only be transferred from $dp[i-1, 2]$.
- When the last round was a jump of 2 steps, the round before last could choose to jump 1 step or 2 steps, that is, $dp[i, 2]$ can be transferred from $dp[i-2, 1]$ or $dp[i-2, 2]$.
As shown in the Figure 14-9 , $dp[i, j]$ represents the number of solutions for state $[i, j]$. At this point, the state transition equation is:
$$
\begin{cases}
dp[i, 1] = dp[i-1, 2] \\
dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
\end{cases}
$$
![Recursive relationship considering constraints](dp_problem_features.assets/climbing_stairs_constraint_state_transfer.png){ class="animation-figure" }
<p align="center"> Figure 14-9 &nbsp; Recursive relationship considering constraints </p>
In the end, returning $dp[n, 1] + dp[n, 2]$ will do, the sum of the two representing the total number of solutions for climbing to the $n$-th step:
=== "Python"
```python title="climbing_stairs_constraint_dp.py"
def climbing_stairs_constraint_dp(n: int) -> int:
"""带约束爬楼梯：动态规划"""
if n == 1 or n == 2:
return 1
# 初始化 dp 表，用于存储子问题的解
dp = [[0] * 3 for _ in range(n + 1)]
# 初始状态：预设最小子问题的解
dp[1][1], dp[1][2] = 1, 0
dp[2][1], dp[2][2] = 0, 1
# 状态转移：从较小子问题逐步求解较大子问题
for i in range(3, n + 1):
dp[i][1] = dp[i - 1][2]
dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
return dp[n][1] + dp[n][2]
```
=== "C++"
```cpp title="climbing_stairs_constraint_dp.cpp"
/* 带约束爬楼梯：动态规划 */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
vector<vector<int>> dp(n + 1, vector<int>(3, 0));
// 初始状态：预设最小子问题的解
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
=== "Java"
```java title="climbing_stairs_constraint_dp.java"
/* 带约束爬楼梯：动态规划 */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
int[][] dp = new int[n + 1][3];
// 初始状态：预设最小子问题的解
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
=== "C#"
```csharp title="climbing_stairs_constraint_dp.cs"
/* 带约束爬楼梯：动态规划 */
int ClimbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
int[,] dp = new int[n + 1, 3];
// 初始状态：预设最小子问题的解
dp[1, 1] = 1;
dp[1, 2] = 0;
dp[2, 1] = 0;
dp[2, 2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i, 1] = dp[i - 1, 2];
dp[i, 2] = dp[i - 2, 1] + dp[i - 2, 2];
}
return dp[n, 1] + dp[n, 2];
}
```
=== "Go"
```go title="climbing_stairs_constraint_dp.go"
/* 带约束爬楼梯：动态规划 */
func climbingStairsConstraintDP(n int) int {
if n == 1 || n == 2 {
return 1
}
// 初始化 dp 表，用于存储子问题的解
dp := make([][3]int, n+1)
// 初始状态：预设最小子问题的解
dp[1][1] = 1
dp[1][2] = 0
dp[2][1] = 0
dp[2][2] = 1
// 状态转移：从较小子问题逐步求解较大子问题
for i := 3; i <= n; i++ {
dp[i][1] = dp[i-1][2]
dp[i][2] = dp[i-2][1] + dp[i-2][2]
}
return dp[n][1] + dp[n][2]
}
```
=== "Swift"
```swift title="climbing_stairs_constraint_dp.swift"
/* 带约束爬楼梯：动态规划 */
func climbingStairsConstraintDP(n: Int) -> Int {
if n == 1 || n == 2 {
return 1
}
// 初始化 dp 表，用于存储子问题的解
var dp = Array(repeating: Array(repeating: 0, count: 3), count: n + 1)
// 初始状态：预设最小子问题的解
dp[1][1] = 1
dp[1][2] = 0
dp[2][1] = 0
dp[2][2] = 1
// 状态转移：从较小子问题逐步求解较大子问题
for i in 3 ... n {
dp[i][1] = dp[i - 1][2]
dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
}
return dp[n][1] + dp[n][2]
}
```
=== "JS"
```javascript title="climbing_stairs_constraint_dp.js"
/* 带约束爬楼梯：动态规划 */
function climbingStairsConstraintDP(n) {
if (n === 1 || n === 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
const dp = Array.from(new Array(n + 1), () => new Array(3));
// 初始状态：预设最小子问题的解
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for (let i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
=== "TS"
```typescript title="climbing_stairs_constraint_dp.ts"
/* 带约束爬楼梯：动态规划 */
function climbingStairsConstraintDP(n: number): number {
if (n === 1 || n === 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
const dp = Array.from({ length: n + 1 }, () => new Array(3));
// 初始状态：预设最小子问题的解
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for (let i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
=== "Dart"
```dart title="climbing_stairs_constraint_dp.dart"
/* 带约束爬楼梯：动态规划 */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
List<List<int>> dp = List.generate(n + 1, (index) => List.filled(3, 0));
// 初始状态：预设最小子问题的解
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
=== "Rust"
```rust title="climbing_stairs_constraint_dp.rs"
/* 带约束爬楼梯：动态规划 */
fn climbing_stairs_constraint_dp(n: usize) -> i32 {
if n == 1 || n == 2 {
return 1;
};
// 初始化 dp 表，用于存储子问题的解
let mut dp = vec![vec![-1; 3]; n + 1];
// 初始状态：预设最小子问题的解
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for i in 3..=n {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
dp[n][1] + dp[n][2]
}
```
=== "C"
```c title="climbing_stairs_constraint_dp.c"
/* 带约束爬楼梯：动态规划 */
int climbingStairsConstraintDP(int n) {
if (n == 1 || n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
int **dp = malloc((n + 1) * sizeof(int *));
for (int i = 0; i <= n; i++) {
dp[i] = calloc(3, sizeof(int));
}
// 初始状态：预设最小子问题的解
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for (int i = 3; i <= n; i++) {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
int res = dp[n][1] + dp[n][2];
// 释放内存
for (int i = 0; i <= n; i++) {
free(dp[i]);
}
free(dp);
return res;
}
```
=== "Kotlin"
```kotlin title="climbing_stairs_constraint_dp.kt"
/* 带约束爬楼梯：动态规划 */
fun climbingStairsConstraintDP(n: Int): Int {
if (n == 1 || n == 2) {
return 1
}
// 初始化 dp 表，用于存储子问题的解
val dp = Array(n + 1) { IntArray(3) }
// 初始状态：预设最小子问题的解
dp[1][1] = 1
dp[1][2] = 0
dp[2][1] = 0
dp[2][2] = 1
// 状态转移：从较小子问题逐步求解较大子问题
for (i in 3..n) {
dp[i][1] = dp[i - 1][2]
dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
}
return dp[n][1] + dp[n][2]
}
```
=== "Ruby"
```ruby title="climbing_stairs_constraint_dp.rb"
[class]{}-[func]{climbing_stairs_constraint_dp}
```
=== "Zig"
```zig title="climbing_stairs_constraint_dp.zig"
// 带约束爬楼梯：动态规划
fn climbingStairsConstraintDP(comptime n: usize) i32 {
if (n == 1 or n == 2) {
return 1;
}
// 初始化 dp 表，用于存储子问题的解
var dp = [_][3]i32{ [_]i32{ -1, -1, -1 } } ** (n + 1);
// 初始状态：预设最小子问题的解
dp[1][1] = 1;
dp[1][2] = 0;
dp[2][1] = 0;
dp[2][2] = 1;
// 状态转移：从较小子问题逐步求解较大子问题
for (3..n + 1) |i| {
dp[i][1] = dp[i - 1][2];
dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
}
return dp[n][1] + dp[n][2];
}
```
??? pythontutor "Code Visualization"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_constraint_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B8%A6%E7%BA%A6%E6%9D%9F%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%203%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D%5B1%5D,%20dp%5B1%5D%5B2%5D%20%3D%201,%200%0A%20%20%20%20dp%5B2%5D%5B1%5D,%20dp%5B2%5D%5B2%5D%20%3D%200,%201%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B1%5D%20%3D%20dp%5Bi%20-%201%5D%5B2%5D%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B2%5D%20%3D%20dp%5Bi%20-%202%5D%5B1%5D%20%2B%20dp%5Bi%20-%202%5D%5B2%5D%0A%20%20%20%20return%20dp%5Bn%5D%5B1%5D%20%2B%20dp%5Bn%5D%5B2%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_constraint_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_constraint_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B8%A6%E7%BA%A6%E6%9D%9F%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%203%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D%5B1%5D,%20dp%5B1%5D%5B2%5D%20%3D%201,%200%0A%20%20%20%20dp%5B2%5D%5B1%5D,%20dp%5B2%5D%5B2%5D%20%3D%200,%201%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B1%5D%20%3D%20dp%5Bi%20-%201%5D%5B2%5D%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B2%5D%20%3D%20dp%5Bi%20-%202%5D%5B1%5D%20%2B%20dp%5Bi%20-%202%5D%5B2%5D%0A%20%20%20%20return%20dp%5Bn%5D%5B1%5D%20%2B%20dp%5Bn%5D%5B2%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_constraint_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
In the above cases, since we only need to consider the previous state, we can still meet the statelessness by expanding the state definition. However, some problems have very serious "state effects".
!!! question "Stair climbing with obstacle generation"
Given a staircase with $n$ steps, you can go up 1 or 2 steps each time. **It is stipulated that when climbing to the $i$-th step, the system automatically places an obstacle on the $2i$-th step, and thereafter all rounds are not allowed to jump to the $2i$-th step**. For example, if the first two rounds jump to the 2nd and 3rd steps, then later you cannot jump to the 4th and 6th steps. How many ways are there to climb to the top?
In this problem, the next jump depends on all past states, as each jump places obstacles on higher steps, affecting future jumps. For such problems, dynamic programming often struggles to solve.
In fact, many complex combinatorial optimization problems (such as the traveling salesman problem) do not satisfy statelessness. For these kinds of problems, we usually choose to use other methods, such as heuristic search, genetic algorithms, reinforcement learning, etc., to obtain usable local optimal solutions within a limited time.
en/docs/chapter_dynamic_programming/dp_solution_pipeline.md
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comments: true
---
# 14.6 &nbsp; Edit distance problem
Edit distance, also known as Levenshtein distance, refers to the minimum number of modifications required to transform one string into another, commonly used in information retrieval and natural language processing to measure the similarity between two sequences.
!!! question
Given two strings $s$ and $t$, return the minimum number of edits required to transform $s$ into $t$.
You can perform three types of edits on a string: insert a character, delete a character, or replace a character with any other character.
As shown in the Figure 14-27 , transforming `kitten` into `sitting` requires 3 edits, including 2 replacements and 1 insertion; transforming `hello` into `algo` requires 3 steps, including 2 replacements and 1 deletion.
![Example data of edit distance](edit_distance_problem.assets/edit_distance_example.png){ class="animation-figure" }
<p align="center"> Figure 14-27 &nbsp; Example data of edit distance </p>
**The edit distance problem can naturally be explained with a decision tree model**. Strings correspond to tree nodes, and a round of decision (an edit operation) corresponds to an edge of the tree.
As shown in the Figure 14-28 , with unrestricted operations, each node can derive many edges, each corresponding to one operation, meaning there are many possible paths to transform `hello` into `algo`.
From the perspective of the decision tree, the goal of this problem is to find the shortest path between the node `hello` and the node `algo`.
![Edit distance problem represented based on decision tree model](edit_distance_problem.assets/edit_distance_decision_tree.png){ class="animation-figure" }
<p align="center"> Figure 14-28 &nbsp; Edit distance problem represented based on decision tree model </p>
### 1. &nbsp; Dynamic programming approach
**Step one: Think about each round of decision, define the state, thus obtaining the $dp$ table**
Each round of decision involves performing one edit operation on string $s$.
We aim to gradually reduce the problem size during the edit process, which enables us to construct subproblems. Let the lengths of strings $s$ and $t$ be $n$ and $m$, respectively. We first consider the tail characters of both strings $s[n-1]$ and $t[m-1]$.
- If $s[n-1]$ and $t[m-1]$ are the same, we can skip them and directly consider $s[n-2]$ and $t[m-2]$.
- If $s[n-1]$ and $t[m-1]$ are different, we need to perform one edit on $s$ (insert, delete, replace) so that the tail characters of the two strings match, allowing us to skip them and consider a smaller-scale problem.
Thus, each round of decision (edit operation) in string $s$ changes the remaining characters in $s$ and $t$ to be matched. Therefore, the state is the $i$-th and $j$-th characters currently considered in $s$ and $t$, denoted as $[i, j]$.
State $[i, j]$ corresponds to the subproblem: **The minimum number of edits required to change the first $i$ characters of $s$ into the first $j$ characters of $t$**.
From this, we obtain a two-dimensional $dp$ table of size $(i+1) \times (j+1)$.
**Step two: Identify the optimal substructure and then derive the state transition equation**
Consider the subproblem $dp[i, j]$, whose corresponding tail characters of the two strings are $s[i-1]$ and $t[j-1]$, which can be divided into three scenarios as shown below.
1. Add $t[j-1]$ after $s[i-1]$, then the remaining subproblem is $dp[i, j-1]$.
2. Delete $s[i-1]$, then the remaining subproblem is $dp[i-1, j]$.
3. Replace $s[i-1]$ with $t[j-1]$, then the remaining subproblem is $dp[i-1, j-1]$.
![State transition of edit distance](edit_distance_problem.assets/edit_distance_state_transfer.png){ class="animation-figure" }
<p align="center"> Figure 14-29 &nbsp; State transition of edit distance </p>
Based on the analysis above, we can determine the optimal substructure: The minimum number of edits for $dp[i, j]$ is the minimum among $dp[i, j-1]$, $dp[i-1, j]$, and $dp[i-1, j-1]$, plus the edit step $1$. The corresponding state transition equation is:
$$
dp[i, j] = \min(dp[i, j-1], dp[i-1, j], dp[i-1, j-1]) + 1
$$
Please note, **when $s[i-1]$ and $t[j-1]$ are the same, no edit is required for the current character**, in which case the state transition equation is:
$$
dp[i, j] = dp[i-1, j-1]
$$
**Step three: Determine the boundary conditions and the order of state transitions**
When both strings are empty, the number of edits is $0$, i.e., $dp[0, 0] = 0$. When $s$ is empty but $t$ is not, the minimum number of edits equals the length of $t$, that is, the first row $dp[0, j] = j$. When $s$ is not empty but $t$ is, the minimum number of edits equals the length of $s$, that is, the first column $dp[i, 0] = i$.
Observing the state transition equation, solving $dp[i, j]$ depends on the solutions to the left, above, and upper left, so a double loop can be used to traverse the entire $dp$ table in the correct order.
### 2. &nbsp; Code implementation
=== "Python"
```python title="edit_distance.py"
def edit_distance_dp(s: str, t: str) -> int:
"""编辑距离：动态规划"""
n, m = len(s), len(t)
dp = [[0] * (m + 1) for _ in range(n + 1)]
# 状态转移：首行首列
for i in range(1, n + 1):
dp[i][0] = i
for j in range(1, m + 1):
dp[0][j] = j
# 状态转移：其余行和列
for i in range(1, n + 1):
for j in range(1, m + 1):
if s[i - 1] == t[j - 1]:
# 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1]
else:
# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1
return dp[n][m]
```
=== "C++"
```cpp title="edit_distance.cpp"
/* 编辑距离：动态规划 */
int editDistanceDP(string s, string t) {
int n = s.length(), m = t.length();
vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
// 状态转移：首行首列
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = min(min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
=== "Java"
```java title="edit_distance.java"
/* 编辑距离：动态规划 */
int editDistanceDP(String s, String t) {
int n = s.length(), m = t.length();
int[][] dp = new int[n + 1][m + 1];
// 状态转移：首行首列
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s.charAt(i - 1) == t.charAt(j - 1)) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
=== "C#"
```csharp title="edit_distance.cs"
/* 编辑距离：动态规划 */
int EditDistanceDP(string s, string t) {
int n = s.Length, m = t.Length;
int[,] dp = new int[n + 1, m + 1];
// 状态转移：首行首列
for (int i = 1; i <= n; i++) {
dp[i, 0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0, j] = j;
}
// 状态转移：其余行和列
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[i, j] = dp[i - 1, j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i, j] = Math.Min(Math.Min(dp[i, j - 1], dp[i - 1, j]), dp[i - 1, j - 1]) + 1;
}
}
}
return dp[n, m];
}
```
=== "Go"
```go title="edit_distance.go"
/* 编辑距离：动态规划 */
func editDistanceDP(s string, t string) int {
n := len(s)
m := len(t)
dp := make([][]int, n+1)
for i := 0; i <= n; i++ {
dp[i] = make([]int, m+1)
}
// 状态转移：首行首列
for i := 1; i <= n; i++ {
dp[i][0] = i
}
for j := 1; j <= m; j++ {
dp[0][j] = j
}
// 状态转移：其余行和列
for i := 1; i <= n; i++ {
for j := 1; j <= m; j++ {
if s[i-1] == t[j-1] {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i-1][j-1]
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = MinInt(MinInt(dp[i][j-1], dp[i-1][j]), dp[i-1][j-1]) + 1
}
}
}
return dp[n][m]
}
```
=== "Swift"
```swift title="edit_distance.swift"
/* 编辑距离：动态规划 */
func editDistanceDP(s: String, t: String) -> Int {
let n = s.utf8CString.count
let m = t.utf8CString.count
var dp = Array(repeating: Array(repeating: 0, count: m + 1), count: n + 1)
// 状态转移：首行首列
for i in 1 ... n {
dp[i][0] = i
}
for j in 1 ... m {
dp[0][j] = j
}
// 状态转移：其余行和列
for i in 1 ... n {
for j in 1 ... m {
if s.utf8CString[i - 1] == t.utf8CString[j - 1] {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1]
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = min(min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1
}
}
}
return dp[n][m]
}
```
=== "JS"
```javascript title="edit_distance.js"
/* 编辑距离：动态规划 */
function editDistanceDP(s, t) {
const n = s.length,
m = t.length;
const dp = Array.from({ length: n + 1 }, () => new Array(m + 1).fill(0));
// 状态转移：首行首列
for (let i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (let j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= m; j++) {
if (s.charAt(i - 1) === t.charAt(j - 1)) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] =
Math.min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
=== "TS"
```typescript title="edit_distance.ts"
/* 编辑距离：动态规划 */
function editDistanceDP(s: string, t: string): number {
const n = s.length,
m = t.length;
const dp = Array.from({ length: n + 1 }, () =>
Array.from({ length: m + 1 }, () => 0)
);
// 状态转移：首行首列
for (let i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (let j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= m; j++) {
if (s.charAt(i - 1) === t.charAt(j - 1)) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] =
Math.min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
=== "Dart"
```dart title="edit_distance.dart"
/* 编辑距离：动态规划 */
int editDistanceDP(String s, String t) {
int n = s.length, m = t.length;
List<List<int>> dp = List.generate(n + 1, (_) => List.filled(m + 1, 0));
// 状态转移：首行首列
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = min(min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
=== "Rust"
```rust title="edit_distance.rs"
/* 编辑距离：动态规划 */
fn edit_distance_dp(s: &str, t: &str) -> i32 {
let (n, m) = (s.len(), t.len());
let mut dp = vec![vec![0; m + 1]; n + 1];
// 状态转移：首行首列
for i in 1..=n {
dp[i][0] = i as i32;
}
for j in 1..m {
dp[0][j] = j as i32;
}
// 状态转移：其余行和列
for i in 1..=n {
for j in 1..=m {
if s.chars().nth(i - 1) == t.chars().nth(j - 1) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] =
std::cmp::min(std::cmp::min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
dp[n][m]
}
```
=== "C"
```c title="edit_distance.c"
/* 编辑距离：动态规划 */
int editDistanceDP(char *s, char *t, int n, int m) {
int **dp = malloc((n + 1) * sizeof(int *));
for (int i = 0; i <= n; i++) {
dp[i] = calloc(m + 1, sizeof(int));
}
// 状态转移：首行首列
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= m; j++) {
dp[0][j] = j;
}
// 状态转移：其余行和列
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = myMin(myMin(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
int res = dp[n][m];
// 释放内存
for (int i = 0; i <= n; i++) {
free(dp[i]);
}
return res;
}
```
=== "Kotlin"
```kotlin title="edit_distance.kt"
/* 编辑距离：动态规划 */
fun editDistanceDP(s: String, t: String): Int {
val n = s.length
val m = t.length
val dp = Array(n + 1) { IntArray(m + 1) }
// 状态转移：首行首列
for (i in 1..n) {
dp[i][0] = i
}
for (j in 1..m) {
dp[0][j] = j
}
// 状态转移：其余行和列
for (i in 1..n) {
for (j in 1..m) {
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1]
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = min(min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1
}
}
}
return dp[n][m]
}
```
=== "Ruby"
```ruby title="edit_distance.rb"
[class]{}-[func]{edit_distance_dp}
```
=== "Zig"
```zig title="edit_distance.zig"
// 编辑距离：动态规划
fn editDistanceDP(comptime s: []const u8, comptime t: []const u8) i32 {
comptime var n = s.len;
comptime var m = t.len;
var dp = [_][m + 1]i32{[_]i32{0} ** (m + 1)} ** (n + 1);
// 状态转移：首行首列
for (1..n + 1) |i| {
dp[i][0] = @intCast(i);
}
for (1..m + 1) |j| {
dp[0][j] = @intCast(j);
}
// 状态转移：其余行和列
for (1..n + 1) |i| {
for (1..m + 1) |j| {
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1];
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[i][j] = @min(@min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
```
??? pythontutor "Code Visualization"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20edit_distance_dp%28s%3A%20str,%20t%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BC%96%E8%BE%91%E8%B7%9D%E7%A6%BB%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28m%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%20i%0A%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Bj%5D%20%3D%20j%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20s%5Bi%20-%201%5D%20%3D%3D%20t%5Bj%20-%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E4%B8%A4%E5%AD%97%E7%AC%A6%E7%9B%B8%E7%AD%89%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%B7%B3%E8%BF%87%E6%AD%A4%E4%B8%A4%E5%AD%97%E7%AC%A6%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%3D%20%E6%8F%92%E5%85%A5%E3%80%81%E5%88%A0%E9%99%A4%E3%80%81%E6%9B%BF%E6%8D%A2%E8%BF%99%E4%B8%89%E7%A7%8D%E6%93%8D%E4%BD%9C%E7%9A%84%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%2B%201%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20min%28dp%5Bi%5D%5Bj%20-%201%5D,%20dp%5Bi%20-%201%5D%5Bj%5D,%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%29%20%2B%201%0A%20%20%20%20return%20dp%5Bn%5D%5Bm%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20s%20%3D%20%22bag%22%0A%20%20%20%20t%20%3D%20%22pack%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20edit_distance_dp%28s,%20t%29%0A%20%20%20%20print%28f%22%E5%B0%86%20%7Bs%7D%20%E6%9B%B4%E6%94%B9%E4%B8%BA%20%7Bt%7D%20%E6%9C%80%E5%B0%91%E9%9C%80%E8%A6%81%E7%BC%96%E8%BE%91%20%7Bres%7D%20%E6%AD%A5%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20edit_distance_dp%28s%3A%20str,%20t%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BC%96%E8%BE%91%E8%B7%9D%E7%A6%BB%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28m%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%20i%0A%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Bj%5D%20%3D%20j%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20s%5Bi%20-%201%5D%20%3D%3D%20t%5Bj%20-%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E4%B8%A4%E5%AD%97%E7%AC%A6%E7%9B%B8%E7%AD%89%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%B7%B3%E8%BF%87%E6%AD%A4%E4%B8%A4%E5%AD%97%E7%AC%A6%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%3D%20%E6%8F%92%E5%85%A5%E3%80%81%E5%88%A0%E9%99%A4%E3%80%81%E6%9B%BF%E6%8D%A2%E8%BF%99%E4%B8%89%E7%A7%8D%E6%93%8D%E4%BD%9C%E7%9A%84%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%2B%201%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20min%28dp%5Bi%5D%5Bj%20-%201%5D,%20dp%5Bi%20-%201%5D%5Bj%5D,%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%29%20%2B%201%0A%20%20%20%20return%20dp%5Bn%5D%5Bm%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20s%20%3D%20%22bag%22%0A%20%20%20%20t%20%3D%20%22pack%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20edit_distance_dp%28s,%20t%29%0A%20%20%20%20print%28f%22%E5%B0%86%20%7Bs%7D%20%E6%9B%B4%E6%94%B9%E4%B8%BA%20%7Bt%7D%20%E6%9C%80%E5%B0%91%E9%9C%80%E8%A6%81%E7%BC%96%E8%BE%91%20%7Bres%7D%20%E6%AD%A5%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
As shown below, the process of state transition in the edit distance problem is very similar to that in the knapsack problem, which can be seen as filling a two-dimensional grid.
=== "<1>"
![Dynamic programming process of edit distance](edit_distance_problem.assets/edit_distance_dp_step1.png){ class="animation-figure" }
=== "<2>"
![edit_distance_dp_step2](edit_distance_problem.assets/edit_distance_dp_step2.png){ class="animation-figure" }
=== "<3>"
![edit_distance_dp_step3](edit_distance_problem.assets/edit_distance_dp_step3.png){ class="animation-figure" }
=== "<4>"
![edit_distance_dp_step4](edit_distance_problem.assets/edit_distance_dp_step4.png){ class="animation-figure" }
=== "<5>"
![edit_distance_dp_step5](edit_distance_problem.assets/edit_distance_dp_step5.png){ class="animation-figure" }
=== "<6>"
![edit_distance_dp_step6](edit_distance_problem.assets/edit_distance_dp_step6.png){ class="animation-figure" }
=== "<7>"
![edit_distance_dp_step7](edit_distance_problem.assets/edit_distance_dp_step7.png){ class="animation-figure" }
=== "<8>"
![edit_distance_dp_step8](edit_distance_problem.assets/edit_distance_dp_step8.png){ class="animation-figure" }
=== "<9>"
![edit_distance_dp_step9](edit_distance_problem.assets/edit_distance_dp_step9.png){ class="animation-figure" }
=== "<10>"
![edit_distance_dp_step10](edit_distance_problem.assets/edit_distance_dp_step10.png){ class="animation-figure" }
=== "<11>"
![edit_distance_dp_step11](edit_distance_problem.assets/edit_distance_dp_step11.png){ class="animation-figure" }
=== "<12>"
![edit_distance_dp_step12](edit_distance_problem.assets/edit_distance_dp_step12.png){ class="animation-figure" }
=== "<13>"
![edit_distance_dp_step13](edit_distance_problem.assets/edit_distance_dp_step13.png){ class="animation-figure" }
=== "<14>"
![edit_distance_dp_step14](edit_distance_problem.assets/edit_distance_dp_step14.png){ class="animation-figure" }
=== "<15>"
![edit_distance_dp_step15](edit_distance_problem.assets/edit_distance_dp_step15.png){ class="animation-figure" }
<p align="center"> Figure 14-30 &nbsp; Dynamic programming process of edit distance </p>
### 3. &nbsp; Space optimization
Since $dp[i, j]$ is derived from the solutions above $dp[i-1, j]$, to the left $dp[i, j-1]$, and to the upper left $dp[i-1, j-1]$, and direct traversal will lose the upper left solution $dp[i-1, j-1]$, and reverse traversal cannot build $dp[i, j-1]$ in advance, therefore, both traversal orders are not feasible.
For this reason, we can use a variable `leftup` to temporarily store the solution from the upper left $dp[i-1, j-1]$, thus only needing to consider the solutions to the left and above. This situation is similar to the unbounded knapsack problem, allowing for direct traversal. The code is as follows:
=== "Python"
```python title="edit_distance.py"
def edit_distance_dp_comp(s: str, t: str) -> int:
"""编辑距离：空间优化后的动态规划"""
n, m = len(s), len(t)
dp = [0] * (m + 1)
# 状态转移：首行
for j in range(1, m + 1):
dp[j] = j
# 状态转移：其余行
for i in range(1, n + 1):
# 状态转移：首列
leftup = dp[0] # 暂存 dp[i-1, j-1]
dp[0] += 1
# 状态转移：其余列
for j in range(1, m + 1):
temp = dp[j]
if s[i - 1] == t[j - 1]:
# 若两字符相等，则直接跳过此两字符
dp[j] = leftup
else:
# 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = min(dp[j - 1], dp[j], leftup) + 1
leftup = temp # 更新为下一轮的 dp[i-1, j-1]
return dp[m]
```
=== "C++"
```cpp title="edit_distance.cpp"
/* 编辑距离：空间优化后的动态规划 */
int editDistanceDPComp(string s, string t) {
int n = s.length(), m = t.length();
vector<int> dp(m + 1, 0);
// 状态转移：首行
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "Java"
```java title="edit_distance.java"
/* 编辑距离：空间优化后的动态规划 */
int editDistanceDPComp(String s, String t) {
int n = s.length(), m = t.length();
int[] dp = new int[m + 1];
// 状态转移：首行
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s.charAt(i - 1) == t.charAt(j - 1)) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = Math.min(Math.min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "C#"
```csharp title="edit_distance.cs"
/* 编辑距离：空间优化后的动态规划 */
int EditDistanceDPComp(string s, string t) {
int n = s.Length, m = t.Length;
int[] dp = new int[m + 1];
// 状态转移：首行
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = Math.Min(Math.Min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "Go"
```go title="edit_distance.go"
/* 编辑距离：空间优化后的动态规划 */
func editDistanceDPComp(s string, t string) int {
n := len(s)
m := len(t)
dp := make([]int, m+1)
// 状态转移：首行
for j := 1; j <= m; j++ {
dp[j] = j
}
// 状态转移：其余行
for i := 1; i <= n; i++ {
// 状态转移：首列
leftUp := dp[0] // 暂存 dp[i-1, j-1]
dp[0] = i
// 状态转移：其余列
for j := 1; j <= m; j++ {
temp := dp[j]
if s[i-1] == t[j-1] {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftUp
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = MinInt(MinInt(dp[j-1], dp[j]), leftUp) + 1
}
leftUp = temp // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m]
}
```
=== "Swift"
```swift title="edit_distance.swift"
/* 编辑距离：空间优化后的动态规划 */
func editDistanceDPComp(s: String, t: String) -> Int {
let n = s.utf8CString.count
let m = t.utf8CString.count
var dp = Array(repeating: 0, count: m + 1)
// 状态转移：首行
for j in 1 ... m {
dp[j] = j
}
// 状态转移：其余行
for i in 1 ... n {
// 状态转移：首列
var leftup = dp[0] // 暂存 dp[i-1, j-1]
dp[0] = i
// 状态转移：其余列
for j in 1 ... m {
let temp = dp[j]
if s.utf8CString[i - 1] == t.utf8CString[j - 1] {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1
}
leftup = temp // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m]
}
```
=== "JS"
```javascript title="edit_distance.js"
/* 编辑距离：状态压缩后的动态规划 */
function editDistanceDPComp(s, t) {
const n = s.length,
m = t.length;
const dp = new Array(m + 1).fill(0);
// 状态转移：首行
for (let j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (let i = 1; i <= n; i++) {
// 状态转移：首列
let leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (let j = 1; j <= m; j++) {
const temp = dp[j];
if (s.charAt(i - 1) === t.charAt(j - 1)) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = Math.min(dp[j - 1], dp[j], leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "TS"
```typescript title="edit_distance.ts"
/* 编辑距离：状态压缩后的动态规划 */
function editDistanceDPComp(s: string, t: string): number {
const n = s.length,
m = t.length;
const dp = new Array(m + 1).fill(0);
// 状态转移：首行
for (let j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (let i = 1; i <= n; i++) {
// 状态转移：首列
let leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (let j = 1; j <= m; j++) {
const temp = dp[j];
if (s.charAt(i - 1) === t.charAt(j - 1)) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = Math.min(dp[j - 1], dp[j], leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "Dart"
```dart title="edit_distance.dart"
/* 编辑距离：空间优化后的动态规划 */
int editDistanceDPComp(String s, String t) {
int n = s.length, m = t.length;
List<int> dp = List.filled(m + 1, 0);
// 状态转移：首行
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m];
}
```
=== "Rust"
```rust title="edit_distance.rs"
/* 编辑距离：空间优化后的动态规划 */
fn edit_distance_dp_comp(s: &str, t: &str) -> i32 {
let (n, m) = (s.len(), t.len());
let mut dp = vec![0; m + 1];
// 状态转移：首行
for j in 1..m {
dp[j] = j as i32;
}
// 状态转移：其余行
for i in 1..=n {
// 状态转移：首列
let mut leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i as i32;
// 状态转移：其余列
for j in 1..=m {
let temp = dp[j];
if s.chars().nth(i - 1) == t.chars().nth(j - 1) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = std::cmp::min(std::cmp::min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
dp[m]
}
```
=== "C"
```c title="edit_distance.c"
/* 编辑距离：空间优化后的动态规划 */
int editDistanceDPComp(char *s, char *t, int n, int m) {
int *dp = calloc(m + 1, sizeof(int));
// 状态转移：首行
for (int j = 1; j <= m; j++) {
dp[j] = j;
}
// 状态转移：其余行
for (int i = 1; i <= n; i++) {
// 状态转移：首列
int leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = i;
// 状态转移：其余列
for (int j = 1; j <= m; j++) {
int temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = myMin(myMin(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
int res = dp[m];
// 释放内存
free(dp);
return res;
}
```
=== "Kotlin"
```kotlin title="edit_distance.kt"
/* 编辑距离：空间优化后的动态规划 */
fun editDistanceDPComp(s: String, t: String): Int {
val n = s.length
val m = t.length
val dp = IntArray(m + 1)
// 状态转移：首行
for (j in 1..m) {
dp[j] = j
}
// 状态转移：其余行
for (i in 1..n) {
// 状态转移：首列
var leftup = dp[0] // 暂存 dp[i-1, j-1]
dp[0] = i
// 状态转移：其余列
for (j in 1..m) {
val temp = dp[j]
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1
}
leftup = temp // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m]
}
```
=== "Ruby"
```ruby title="edit_distance.rb"
[class]{}-[func]{edit_distance_dp_comp}
```
=== "Zig"
```zig title="edit_distance.zig"
// 编辑距离：空间优化后的动态规划
fn editDistanceDPComp(comptime s: []const u8, comptime t: []const u8) i32 {
comptime var n = s.len;
comptime var m = t.len;
var dp = [_]i32{0} ** (m + 1);
// 状态转移：首行
for (1..m + 1) |j| {
dp[j] = @intCast(j);
}
// 状态转移：其余行
for (1..n + 1) |i| {
// 状态转移：首列
var leftup = dp[0]; // 暂存 dp[i-1, j-1]
dp[0] = @intCast(i);
// 状态转移：其余列
for (1..m + 1) |j| {
var temp = dp[j];
if (s[i - 1] == t[j - 1]) {
// 若两字符相等，则直接跳过此两字符
dp[j] = leftup;
} else {
// 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
dp[j] = @min(@min(dp[j - 1], dp[j]), leftup) + 1;
}
leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
}
}
return dp[m];
}
```
??? pythontutor "Code Visualization"
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# Chapter 14. &nbsp; Dynamic programming
![Dynamic programming](../assets/covers/chapter_dynamic_programming.jpg){ class="cover-image" }
!!! abstract
Streams merge into rivers, and rivers merge into the sea.
Dynamic programming combines the solutions of small problems to solve bigger problems, step by step leading us to the solution.
## Chapter contents
- [14.1 &nbsp; Introduction to dynamic programming](intro_to_dynamic_programming.md)
- [14.2 &nbsp; Characteristics of DP problems](dp_problem_features.md)
- [14.3 &nbsp; DP problem-solving approach¶](dp_solution_pipeline.md)
- [14.4 &nbsp; 0-1 Knapsack problem](knapsack_problem.md)
- [14.5 &nbsp; Unbounded knapsack problem](unbounded_knapsack_problem.md)
- [14.6 &nbsp; Edit distance problem](edit_distance_problem.md)
- [14.7 &nbsp; Summary](summary.md)
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# 14.7 &nbsp; Summary
- Dynamic programming decomposes problems and improves computational efficiency by avoiding redundant computations through storing solutions of subproblems.
- Without considering time, all dynamic programming problems can be solved using backtracking (brute force search), but the recursion tree has many overlapping subproblems, resulting in very low efficiency. By introducing a memorization list, it's possible to store solutions of all computed subproblems, ensuring that overlapping subproblems are only computed once.
- Memorization search is a top-down recursive solution, whereas dynamic programming corresponds to a bottom-up iterative approach, akin to "filling out a table." Since the current state only depends on certain local states, we can eliminate one dimension of the dp table to reduce space complexity.
- Decomposition of subproblems is a universal algorithmic approach, differing in characteristics among divide and conquer, dynamic programming, and backtracking.
- Dynamic programming problems have three main characteristics: overlapping subproblems, optimal substructure, and no aftereffects.
- If the optimal solution of the original problem can be constructed from the optimal solutions of its subproblems, it has an optimal substructure.
- No aftereffects mean that the future development of a state depends only on the current state and not on all past states experienced. Many combinatorial optimization problems do not have this property and cannot be quickly solved using dynamic programming.
**Knapsack problem**
- The knapsack problem is one of the most typical dynamic programming problems, with variants including the 0-1 knapsack, unbounded knapsack, and multiple knapsacks.
- The state definition of the 0-1 knapsack is the maximum value in a knapsack of capacity $c$ with the first $i$ items. Based on decisions not to include or to include an item in the knapsack, optimal substructures can be identified and state transition equations constructed. In space optimization, since each state depends on the state directly above and to the upper left, the list should be traversed in reverse order to avoid overwriting the upper left state.
- In the unbounded knapsack problem, there is no limit on the number of each kind of item that can be chosen, thus the state transition for including items differs from the 0-1 knapsack. Since the state depends on the state directly above and to the left, space optimization should involve forward traversal.
- The coin change problem is a variant of the unbounded knapsack problem, shifting from seeking the “maximum” value to seeking the “minimum” number of coins, thus the state transition equation should change $\max()$ to $\min()$. From pursuing “not exceeding” the capacity of the knapsack to seeking exactly the target amount, thus use $amt + 1$ to represent the invalid solution of “unable to make up the target amount.”
- Coin Change Problem II shifts from seeking the “minimum number of coins” to seeking the “number of coin combinations,” changing the state transition equation accordingly from $\min()$ to summation operator.
**Edit distance problem**
- Edit distance (Levenshtein distance) measures the similarity between two strings, defined as the minimum number of editing steps needed to change one string into another, with editing operations including adding, deleting, or replacing.
- The state definition for the edit distance problem is the minimum number of editing steps needed to change the first $i$ characters of $s$ into the first $j$ characters of $t$. When $s[i] \ne t[j]$, there are three decisions: add, delete, replace, each with their corresponding residual subproblems. From this, optimal substructures can be identified, and state transition equations built. When $s[i] = t[j]$, no editing of the current character is necessary.
- In edit distance, the state depends on the state directly above, to the left, and to the upper left. Therefore, after space optimization, neither forward nor reverse traversal can correctly perform state transitions. To address this, we use a variable to temporarily store the upper left state, making it equivalent to the situation in the unbounded knapsack problem, allowing for forward traversal after space optimization.
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