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---
comments: true
---
# 13.4   N-Queens Problem
!!! question
According to the rules of chess, a queen can attack any piece in the same row, column, or diagonal. Given $n$ queens and an $n \times n$ chessboard, find an arrangement such that no two queens can attack each other.
As shown in Figure 13-15, when $n = 4$, there are two solutions that can be found. From the perspective of the backtracking algorithm, an $n \times n$ chessboard has $n^2$ squares, which provide all the choices `choices`. During the process of placing queens one by one, the chessboard state changes continuously, and the chessboard at each moment represents the state `state`.
![Solution to the 4-queens problem](n_queens_problem.assets/solution_4_queens.png){ class="animation-figure" }
<p align="center"> Figure 13-15 &nbsp; Solution to the 4-queens problem </p>
Figure 13-16 illustrates the three constraints of this problem: **multiple queens cannot be in the same row, the same column, or on the same diagonal**. It is worth noting that diagonals are divided into two types: the main diagonal `\` and the anti-diagonal `/`.
![Constraints of the n-queens problem](n_queens_problem.assets/n_queens_constraints.png){ class="animation-figure" }
<p align="center"> Figure 13-16 &nbsp; Constraints of the n-queens problem </p>
### 1. &nbsp; Row-By-Row Placement Strategy
Since both the number of queens and the number of rows on the chessboard are $n$, we can easily derive a conclusion: **each row of the chessboard allows one and only one queen to be placed**.
This means we can adopt a row-by-row placement strategy: starting from the first row, place one queen in each row until the last row is completed.
Figure 13-17 shows the row-by-row placement process for the 4-queens problem. Due to space limitations, the figure only expands one search branch of the first row, and all schemes that violate the column or diagonal constraints are pruned.
![Row-by-row placement strategy](n_queens_problem.assets/n_queens_placing.png){ class="animation-figure" }
<p align="center"> Figure 13-17 &nbsp; Row-by-row placement strategy </p>
Essentially, **the row-by-row placement strategy serves a pruning function**, as it avoids all search branches where multiple queens appear in the same row.
### 2. &nbsp; Column and Diagonal Pruning
To satisfy the column constraint, we can use a boolean array `cols` of length $n$ to record whether each column has a queen. Before each placement decision, we use `cols` to prune columns that already have queens, and dynamically update the state of `cols` during backtracking.
!!! tip
Please note that the origin of the matrix is located in the upper-left corner, where the row index increases from top to bottom, and the column index increases from left to right.
So how do we handle diagonal constraints? Consider a square on the chessboard with row and column indices $(row, col)$. If we select a specific main diagonal in the matrix, we find that all squares on that diagonal have the same difference between their row and column indices, **meaning that $row - col$ is a constant value for all squares on the main diagonal**.
In other words, if two squares satisfy $row_1 - col_1 = row_2 - col_2$, they must be on the same main diagonal. Using this pattern, we can use the array `diags1` shown in Figure 13-18 to record whether there is a queen on each main diagonal.
Similarly, **for all squares on an anti-diagonal, the sum $row + col$ is a constant value**. We can likewise use the array `diags2` to handle anti-diagonal constraints.
![Handling column and diagonal constraints](n_queens_problem.assets/n_queens_cols_diagonals.png){ class="animation-figure" }
<p align="center"> Figure 13-18 &nbsp; Handling column and diagonal constraints </p>
### 3. &nbsp; Code Implementation
Please note that in an $n \times n$ square matrix, the range of $row - col$ is $[-n + 1, n - 1]$, and the range of $row + col$ is $[0, 2n - 2]$. Therefore, the number of both main diagonals and anti-diagonals is $2n - 1$, meaning the length of both arrays `diags1` and `diags2` is $2n - 1$.
=== "Python"
```python title="n_queens.py"
def backtrack(
row: int,
n: int,
state: list[list[str]],
res: list[list[list[str]]],
cols: list[bool],
diags1: list[bool],
diags2: list[bool],
):
"""Backtracking algorithm: N queens"""
# When all rows are placed, record the solution
if row == n:
res.append([list(row) for row in state])
return
# Traverse all columns
for col in range(n):
# Calculate the main diagonal and anti-diagonal corresponding to this cell
diag1 = row - col + n - 1
diag2 = row + col
# Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if not cols[col] and not diags1[diag1] and not diags2[diag2]:
# Attempt: place the queen in this cell
state[row][col] = "Q"
cols[col] = diags1[diag1] = diags2[diag2] = True
# Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2)
# Backtrack: restore this cell to an empty cell
state[row][col] = "#"
cols[col] = diags1[diag1] = diags2[diag2] = False
def n_queens(n: int) -> list[list[list[str]]]:
"""Solve N queens"""
# Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
state = [["#" for _ in range(n)] for _ in range(n)]
cols = [False] * n # Record whether there is a queen in the column
diags1 = [False] * (2 * n - 1) # Record whether there is a queen on the main diagonal
diags2 = [False] * (2 * n - 1) # Record whether there is a queen on the anti-diagonal
res = []
backtrack(0, n, state, res, cols, diags1, diags2)
return res
```
=== "C++"
```cpp title="n_queens.cpp"
/* Backtracking algorithm: N queens */
void backtrack(int row, int n, vector<vector<string>> &state, vector<vector<vector<string>>> &res, vector<bool> &cols,
vector<bool> &diags1, vector<bool> &diags2) {
// When all rows are placed, record the solution
if (row == n) {
res.push_back(state);
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = "Q";
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#";
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
vector<vector<vector<string>>> nQueens(int n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
vector<vector<string>> state(n, vector<string>(n, "#"));
vector<bool> cols(n, false); // Record whether there is a queen in the column
vector<bool> diags1(2 * n - 1, false); // Record whether there is a queen on the main diagonal
vector<bool> diags2(2 * n - 1, false); // Record whether there is a queen on the anti-diagonal
vector<vector<vector<string>>> res;
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Java"
```java title="n_queens.java"
/* Backtracking algorithm: N queens */
void backtrack(int row, int n, List<List<String>> state, List<List<List<String>>> res,
boolean[] cols, boolean[] diags1, boolean[] diags2) {
// When all rows are placed, record the solution
if (row == n) {
List<List<String>> copyState = new ArrayList<>();
for (List<String> sRow : state) {
copyState.add(new ArrayList<>(sRow));
}
res.add(copyState);
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state.get(row).set(col, "Q");
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state.get(row).set(col, "#");
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
List<List<List<String>>> nQueens(int n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
List<List<String>> state = new ArrayList<>();
for (int i = 0; i < n; i++) {
List<String> row = new ArrayList<>();
for (int j = 0; j < n; j++) {
row.add("#");
}
state.add(row);
}
boolean[] cols = new boolean[n]; // Record whether there is a queen in the column
boolean[] diags1 = new boolean[2 * n - 1]; // Record whether there is a queen on the main diagonal
boolean[] diags2 = new boolean[2 * n - 1]; // Record whether there is a queen on the anti-diagonal
List<List<List<String>>> res = new ArrayList<>();
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "C#"
```csharp title="n_queens.cs"
/* Backtracking algorithm: N queens */
void Backtrack(int row, int n, List<List<string>> state, List<List<List<string>>> res,
bool[] cols, bool[] diags1, bool[] diags2) {
// When all rows are placed, record the solution
if (row == n) {
List<List<string>> copyState = [];
foreach (List<string> sRow in state) {
copyState.Add(new List<string>(sRow));
}
res.Add(copyState);
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = "Q";
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
Backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#";
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
List<List<List<string>>> NQueens(int n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
List<List<string>> state = [];
for (int i = 0; i < n; i++) {
List<string> row = [];
for (int j = 0; j < n; j++) {
row.Add("#");
}
state.Add(row);
}
bool[] cols = new bool[n]; // Record whether there is a queen in the column
bool[] diags1 = new bool[2 * n - 1]; // Record whether there is a queen on the main diagonal
bool[] diags2 = new bool[2 * n - 1]; // Record whether there is a queen on the anti-diagonal
List<List<List<string>>> res = [];
Backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Go"
```go title="n_queens.go"
/* Backtracking algorithm: N queens */
func backtrack(row, n int, state *[][]string, res *[][][]string, cols, diags1, diags2 *[]bool) {
// When all rows are placed, record the solution
if row == n {
newState := make([][]string, len(*state))
for i, _ := range newState {
newState[i] = make([]string, len((*state)[0]))
copy(newState[i], (*state)[i])
}
*res = append(*res, newState)
return
}
// Traverse all columns
for col := 0; col < n; col++ {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
diag1 := row - col + n - 1
diag2 := row + col
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !(*cols)[col] && !(*diags1)[diag1] && !(*diags2)[diag2] {
// Attempt: place the queen in this cell
(*state)[row][col] = "Q"
(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = true, true, true
// Place the next row
backtrack(row+1, n, state, res, cols, diags1, diags2)
// Backtrack: restore this cell to an empty cell
(*state)[row][col] = "#"
(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = false, false, false
}
}
}
/* Solve N queens */
func nQueens(n int) [][][]string {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
state := make([][]string, n)
for i := 0; i < n; i++ {
row := make([]string, n)
for i := 0; i < n; i++ {
row[i] = "#"
}
state[i] = row
}
// Record whether there is a queen in the column
cols := make([]bool, n)
diags1 := make([]bool, 2*n-1)
diags2 := make([]bool, 2*n-1)
res := make([][][]string, 0)
backtrack(0, n, &state, &res, &cols, &diags1, &diags2)
return res
}
```
=== "Swift"
```swift title="n_queens.swift"
/* Backtracking algorithm: N queens */
func backtrack(row: Int, n: Int, state: inout [[String]], res: inout [[[String]]], cols: inout [Bool], diags1: inout [Bool], diags2: inout [Bool]) {
// When all rows are placed, record the solution
if row == n {
res.append(state)
return
}
// Traverse all columns
for col in 0 ..< n {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
let diag1 = row - col + n - 1
let diag2 = row + col
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !cols[col] && !diags1[diag1] && !diags2[diag2] {
// Attempt: place the queen in this cell
state[row][col] = "Q"
cols[col] = true
diags1[diag1] = true
diags2[diag2] = true
// Place the next row
backtrack(row: row + 1, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
// Backtrack: restore this cell to an empty cell
state[row][col] = "#"
cols[col] = false
diags1[diag1] = false
diags2[diag2] = false
}
}
}
/* Solve N queens */
func nQueens(n: Int) -> [[[String]]] {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
var state = Array(repeating: Array(repeating: "#", count: n), count: n)
var cols = Array(repeating: false, count: n) // Record whether there is a queen in the column
var diags1 = Array(repeating: false, count: 2 * n - 1) // Record whether there is a queen on the main diagonal
var diags2 = Array(repeating: false, count: 2 * n - 1) // Record whether there is a queen on the anti-diagonal
var res: [[[String]]] = []
backtrack(row: 0, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
return res
}
```
=== "JS"
```javascript title="n_queens.js"
/* Backtracking algorithm: N queens */
function backtrack(row, n, state, res, cols, diags1, diags2) {
// When all rows are placed, record the solution
if (row === n) {
res.push(state.map((row) => row.slice()));
return;
}
// Traverse all columns
for (let col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
const diag1 = row - col + n - 1;
const diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
function nQueens(n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
const state = Array.from({ length: n }, () => Array(n).fill('#'));
const cols = Array(n).fill(false); // Record whether there is a queen in the column
const diags1 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the main diagonal
const diags2 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the anti-diagonal
const res = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "TS"
```typescript title="n_queens.ts"
/* Backtracking algorithm: N queens */
function backtrack(
row: number,
n: number,
state: string[][],
res: string[][][],
cols: boolean[],
diags1: boolean[],
diags2: boolean[]
): void {
// When all rows are placed, record the solution
if (row === n) {
res.push(state.map((row) => row.slice()));
return;
}
// Traverse all columns
for (let col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
const diag1 = row - col + n - 1;
const diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
function nQueens(n: number): string[][][] {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
const state = Array.from({ length: n }, () => Array(n).fill('#'));
const cols = Array(n).fill(false); // Record whether there is a queen in the column
const diags1 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the main diagonal
const diags2 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the anti-diagonal
const res: string[][][] = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Dart"
```dart title="n_queens.dart"
/* Backtracking algorithm: N queens */
void backtrack(
int row,
int n,
List<List<String>> state,
List<List<List<String>>> res,
List<bool> cols,
List<bool> diags1,
List<bool> diags2,
) {
// When all rows are placed, record the solution
if (row == n) {
List<List<String>> copyState = [];
for (List<String> sRow in state) {
copyState.add(List.from(sRow));
}
res.add(copyState);
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = "Q";
cols[col] = true;
diags1[diag1] = true;
diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#";
cols[col] = false;
diags1[diag1] = false;
diags2[diag2] = false;
}
}
}
/* Solve N queens */
List<List<List<String>>> nQueens(int n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
List<List<String>> state = List.generate(n, (index) => List.filled(n, "#"));
List<bool> cols = List.filled(n, false); // Record whether there is a queen in the column
List<bool> diags1 = List.filled(2 * n - 1, false); // Record whether there is a queen on the main diagonal
List<bool> diags2 = List.filled(2 * n - 1, false); // Record whether there is a queen on the anti-diagonal
List<List<List<String>>> res = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Rust"
```rust title="n_queens.rs"
/* Backtracking algorithm: N queens */
fn backtrack(
row: usize,
n: usize,
state: &mut Vec<Vec<String>>,
res: &mut Vec<Vec<Vec<String>>>,
cols: &mut [bool],
diags1: &mut [bool],
diags2: &mut [bool],
) {
// When all rows are placed, record the solution
if row == n {
res.push(state.clone());
return;
}
// Traverse all columns
for col in 0..n {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
let diag1 = row + n - 1 - col;
let diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !cols[col] && !diags1[diag1] && !diags2[diag2] {
// Attempt: place the queen in this cell
state[row][col] = "Q".into();
(cols[col], diags1[diag1], diags2[diag2]) = (true, true, true);
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#".into();
(cols[col], diags1[diag1], diags2[diag2]) = (false, false, false);
}
}
}
/* Solve N queens */
fn n_queens(n: usize) -> Vec<Vec<Vec<String>>> {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
let mut state: Vec<Vec<String>> = vec![vec!["#".to_string(); n]; n];
let mut cols = vec![false; n]; // Record whether there is a queen in the column
let mut diags1 = vec![false; 2 * n - 1]; // Record whether there is a queen on the main diagonal
let mut diags2 = vec![false; 2 * n - 1]; // Record whether there is a queen on the anti-diagonal
let mut res: Vec<Vec<Vec<String>>> = Vec::new();
backtrack(
0,
n,
&mut state,
&mut res,
&mut cols,
&mut diags1,
&mut diags2,
);
res
}
```
=== "C"
```c title="n_queens.c"
/* Backtracking algorithm: N queens */
void backtrack(int row, int n, char state[MAX_SIZE][MAX_SIZE], char ***res, int *resSize, bool cols[MAX_SIZE],
bool diags1[2 * MAX_SIZE - 1], bool diags2[2 * MAX_SIZE - 1]) {
// When all rows are placed, record the solution
if (row == n) {
res[*resSize] = (char **)malloc(sizeof(char *) * n);
for (int i = 0; i < n; ++i) {
res[*resSize][i] = (char *)malloc(sizeof(char) * (n + 1));
strcpy(res[*resSize][i], state[i]);
}
(*resSize)++;
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, resSize, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
char ***nQueens(int n, int *returnSize) {
char state[MAX_SIZE][MAX_SIZE];
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
state[i][j] = '#';
}
state[i][n] = '\0';
}
bool cols[MAX_SIZE] = {false}; // Record whether there is a queen in the column
bool diags1[2 * MAX_SIZE - 1] = {false}; // Record whether there is a queen on the main diagonal
bool diags2[2 * MAX_SIZE - 1] = {false}; // Record whether there is a queen on the anti-diagonal
char ***res = (char ***)malloc(sizeof(char **) * MAX_SIZE);
*returnSize = 0;
backtrack(0, n, state, res, returnSize, cols, diags1, diags2);
return res;
}
```
=== "Kotlin"
```kotlin title="n_queens.kt"
/* Backtracking algorithm: N queens */
fun backtrack(
row: Int,
n: Int,
state: MutableList<MutableList<String>>,
res: MutableList<MutableList<MutableList<String>>?>,
cols: BooleanArray,
diags1: BooleanArray,
diags2: BooleanArray
) {
// When all rows are placed, record the solution
if (row == n) {
val copyState = mutableListOf<MutableList<String>>()
for (sRow in state) {
copyState.add(sRow.toMutableList())
}
res.add(copyState)
return
}
// Traverse all columns
for (col in 0..<n) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
val diag1 = row - col + n - 1
val diag2 = row + col
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = "Q"
diags2[diag2] = true
diags1[diag1] = diags2[diag2]
cols[col] = diags1[diag1]
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2)
// Backtrack: restore this cell to an empty cell
state[row][col] = "#"
diags2[diag2] = false
diags1[diag1] = diags2[diag2]
cols[col] = diags1[diag1]
}
}
}
/* Solve N queens */
fun nQueens(n: Int): MutableList<MutableList<MutableList<String>>?> {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
val state = mutableListOf<MutableList<String>>()
for (i in 0..<n) {
val row = mutableListOf<String>()
for (j in 0..<n) {
row.add("#")
}
state.add(row)
}
val cols = BooleanArray(n) // Record whether there is a queen in the column
val diags1 = BooleanArray(2 * n - 1) // Record whether there is a queen on the main diagonal
val diags2 = BooleanArray(2 * n - 1) // Record whether there is a queen on the anti-diagonal
val res = mutableListOf<MutableList<MutableList<String>>?>()
backtrack(0, n, state, res, cols, diags1, diags2)
return res
}
```
=== "Ruby"
```ruby title="n_queens.rb"
### Backtracking: n queens ###
def backtrack(row, n, state, res, cols, diags1, diags2)
# When all rows are placed, record the solution
if row == n
res << state.map { |row| row.dup }
return
end
# Traverse all columns
for col in 0...n
# Calculate the main diagonal and anti-diagonal corresponding to this cell
diag1 = row - col + n - 1
diag2 = row + col
# Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !cols[col] && !diags1[diag1] && !diags2[diag2]
# Attempt: place the queen in this cell
state[row][col] = "Q"
cols[col] = diags1[diag1] = diags2[diag2] = true
# Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2)
# Backtrack: restore this cell to an empty cell
state[row][col] = "#"
cols[col] = diags1[diag1] = diags2[diag2] = false
end
end
end
### Solve n queens ###
def n_queens(n)
# Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
state = Array.new(n) { Array.new(n, "#") }
cols = Array.new(n, false) # Record whether there is a queen in the column
diags1 = Array.new(2 * n - 1, false) # Record whether there is a queen on the main diagonal
diags2 = Array.new(2 * n - 1, false) # Record whether there is a queen on the anti-diagonal
res = []
backtrack(0, n, state, res, cols, diags1, diags2)
res
end
```
Placing $n$ queens row by row, considering the column constraint, from the first row to the last row there are $n$, $n-1$, $\dots$, $2$, $1$ choices, using $O(n!)$ time. When recording a solution, it is necessary to copy the matrix `state` and add it to `res`, and the copy operation uses $O(n^2)$ time. Therefore, **the overall time complexity is $O(n! \cdot n^2)$**. In practice, pruning based on diagonal constraints can also significantly reduce the search space, so the search efficiency is often better than the time complexity mentioned above.
The array `state` uses $O(n^2)$ space, and the arrays `cols`, `diags1`, and `diags2` each use $O(n)$ space. The maximum recursion depth is $n$, using $O(n)$ stack frame space. Therefore, **the space complexity is $O(n^2)$**.
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