N-Queens

Jun 26, 2026madhavcodes25

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Description:

The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle. You may return the answer in any order.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space, respectively.

Example 1:

Input: n = 4 Output: [[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]] Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.

Example 2:

Input: n = 1 Output: [["Q"]]

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Approaches:

1. Backtracking

The most effective way to solve the N-Queens problem is using Backtracking. We place queens row by row. For each row, we try to place a queen in each column. We must ensure that the new queen is not under attack by any previously placed queens.

1. Maintain three sets to track the columns and the two diagonals (positive and negative) that are currently occupied by queens.
2. The positive diagonal has a constant property where row + col is the same for all elements on that diagonal.
3. The negative diagonal has a constant property where row - col is the same for all elements on that diagonal.
4. Create a recursive backtrack function that starts at row = 0.
5. If row == n, a valid board configuration has been found, so we format it and add it to our results.
6. For the current row, iterate through each column c. If placing a queen at (row, c) violates any of our sets, skip it.
7. Otherwise, place the queen, update the sets, and recursively call backtrack(row + 1).
8. After exploring that path, remove the queen and back out of the sets to explore other possibilities.

Board Size (N):

Warning: Values above 8 may cause performance issues.

Step 1 of 0:

Solutions Found

0

• Time Complexity: $O(N!)$ where $N$ is the number of queens. For the first row we have $N$ choices, for the next roughly $N-1$, and so on, leading to a factorial time complexity.
• Space Complexity: $O(N^2)$ for storing the board state and the output array, plus $O(N)$ for the recursion stack and the sets used to track attacks.

Solutions:

C++

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        vector<vector<string>> res;
        vector<string> board(n, string(n, '.'));
        vector<int> col(n, 0), posDiag(2 * n, 0), negDiag(2 * n, 0);
        backtrack(0, n, board, res, col, posDiag, negDiag);
        return res;
    }

private:
    void backtrack(int r, int n, vector<string>& board, vector<vector<string>>& res, vector<int>& col, vector<int>& posDiag, vector<int>& negDiag) {
        if (r == n) {
            res.push_back(board);
            return;
        }
        for (int c = 0; c < n; c++) {
            if (col[c] || posDiag[r + c] || negDiag[r - c + n]) continue;
            
            board[r][c] = 'Q';
            col[c] = posDiag[r + c] = negDiag[r - c + n] = 1;
            
            backtrack(r + 1, n, board, res, col, posDiag, negDiag);
            
            board[r][c] = '.';
            col[c] = posDiag[r + c] = negDiag[r - c + n] = 0;
        }
    }
};

Java

class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> res = new ArrayList<>();
        char[][] board = new char[n][n];
        for (int i = 0; i < n; i++) {
            Arrays.fill(board[i], '.');
        }
        
        Set<Integer> col = new HashSet<>();
        Set<Integer> posDiag = new HashSet<>();
        Set<Integer> negDiag = new HashSet<>();
        
        backtrack(0, n, board, res, col, posDiag, negDiag);
        return res;
    }
    
    private void backtrack(int r, int n, char[][] board, List<List<String>> res, Set<Integer> col, Set<Integer> posDiag, Set<Integer> negDiag) {
        if (r == n) {
            List<String> copy = new ArrayList<>();
            for (char[] row : board) {
                copy.add(new String(row));
            }
            res.add(copy);
            return;
        }
        
        for (int c = 0; c < n; c++) {
            if (col.contains(c) || posDiag.contains(r + c) || negDiag.contains(r - c)) {
                continue;
            }
            
            col.add(c);
            posDiag.add(r + c);
            negDiag.add(r - c);
            board[r][c] = 'Q';
            
            backtrack(r + 1, n, board, res, col, posDiag, negDiag);
            
            col.remove(c);
            posDiag.remove(r + c);
            negDiag.remove(r - c);
            board[r][c] = '.';
        }
    }
}

Python

class Solution:
    def solveNQueens(self, n: int) -> List[List[str]]:
        col = set()
        posDiag = set() # (r + c)
        negDiag = set() # (r - c)
        
        res = []
        board = [["."] * n for _ in range(n)]
        
        def backtrack(r):
            if r == n:
                copy = ["".join(row) for row in board]
                res.append(copy)
                return
            
            for c in range(n):
                if c in col or (r + c) in posDiag or (r - c) in negDiag:
                    continue
                
                col.add(c)
                posDiag.add(r + c)
                negDiag.add(r - c)
                board[r][c] = "Q"
                
                backtrack(r + 1)
                
                col.remove(c)
                posDiag.remove(r + c)
                negDiag.remove(r - c)
                board[r][c] = "."
                
        backtrack(0)
        return res

JavaScript

/**
 * @param {number} n
 * @return {string[][]}
 */
const solveNQueens = function(n) {
    const res = [];
    const board = Array.from({ length: n }, () => Array(n).fill('.'));
    const col = new Set();
    const posDiag = new Set();
    const negDiag = new Set();

    const backtrack = (r) => {
        if (r === n) {
            res.push(board.map(row => row.join('')));
            return;
        }

        for (let c = 0; c < n; c++) {
            if (col.has(c) || posDiag.has(r + c) || negDiag.has(r - c)) {
                continue;
            }

            col.add(c);
            posDiag.add(r + c);
            negDiag.add(r - c);
            board[r][c] = 'Q';

            backtrack(r + 1);

            col.delete(c);
            posDiag.delete(r + c);
            negDiag.delete(r - c);
            board[r][c] = '.';
        }
    };

    backtrack(0);
    return res;
};