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# N-Queens LeetCode Solution – Queslers

## Problem – N-Queens LeetCode Solution

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"]]``````

Constraints:

• `1 <= n <= 9`

## N-Queens LeetCode Solution in C++

``````class Solution {
public:
std::vector<std::vector<std::string> > solveNQueens(int n) {
std::vector<std::vector<std::string> > res;
std::vector<std::string> nQueens(n, std::string(n, '.'));
solveNQueens(res, nQueens, 0, n);
return res;
}
private:
void solveNQueens(std::vector<std::vector<std::string> > &res, std::vector<std::string> &nQueens, int row, int &n) {
if (row == n) {
res.push_back(nQueens);
return;
}
for (int col = 0; col != n; ++col)
if (isValid(nQueens, row, col, n)) {
nQueens[row][col] = 'Q';
solveNQueens(res, nQueens, row + 1, n);
nQueens[row][col] = '.';
}
}
bool isValid(std::vector<std::string> &nQueens, int row, int col, int &n) {
//check if the column had a queen before.
for (int i = 0; i != row; ++i)
if (nQueens[i][col] == 'Q')
return false;
//check if the 45° diagonal had a queen before.
for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; --i, --j)
if (nQueens[i][j] == 'Q')
return false;
//check if the 135° diagonal had a queen before.
for (int i = row - 1, j = col + 1; i >= 0 && j < n; --i, ++j)
if (nQueens[i][j] == 'Q')
return false;
return true;
}
};
``````

## N-Queens LeetCode Solution in Python

``````def solveNQueens(self, n):
def DFS(queens, xy_dif, xy_sum):
p = len(queens)
if p==n:
result.append(queens)
return None
for q in range(n):
if q not in queens and p-q not in xy_dif and p+q not in xy_sum:
DFS(queens+[q], xy_dif+[p-q], xy_sum+[p+q])
result = []
DFS([],[],[])
return [ ["."*i + "Q" + "."*(n-i-1) for i in sol] for sol in result]
``````

## N-Queens LeetCode Solution in Java

``````public class Solution {
public List<List<String>> solveNQueens(int n) {
char[][] board = new char[n][n];
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
board[i][j] = '.';
List<List<String>> res = new ArrayList<List<String>>();
dfs(board, 0, res);
return res;
}

private void dfs(char[][] board, int colIndex, List<List<String>> res) {
if(colIndex == board.length) {
return;
}

for(int i = 0; i < board.length; i++) {
if(validate(board, i, colIndex)) {
board[i][colIndex] = 'Q';
dfs(board, colIndex + 1, res);
board[i][colIndex] = '.';
}
}
}

private boolean validate(char[][] board, int x, int y) {
for(int i = 0; i < board.length; i++) {
for(int j = 0; j < y; j++) {
if(board[i][j] == 'Q' && (x + j == y + i || x + y == i + j || x == i))
return false;
}
}

return true;
}

private List<String> construct(char[][] board) {
List<String> res = new LinkedList<String>();
for(int i = 0; i < board.length; i++) {
String s = new String(board[i]);
}
return res;
}
}
``````
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