> For the complete documentation index, see [llms.txt](https://soumyajit4419.gitbook.io/ds-algo/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://soumyajit4419.gitbook.io/ds-algo/recursion-and-backtracking/10.-n-queens.md).

# 12. N-Queens

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***.

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:**

![](https://assets.leetcode.com/uploads/2020/11/13/queens.jpg)

```
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"]]
```

## Solution: (**Backtracking**)

**Approach:**\
**Try each possible cell and check if it is safe or not**\
**check for row, col, diagonal**\
**Diagonal Formula = row + col and row - col**

```cpp
class Solution
{
public:
    vector<vector<string>> res;
    bool isSafe(int row, int col, int n, vector<vector<int>> &v)
    {
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (v[i][j] == 1)
                {
                    if (i == row || j == col || row + col == i + j || row - col == i - j)
                    {
                        return false;
                    }
                }
            }
        }
        return true;
    }

    void findPos(int row, int n, vector<vector<int>> &v)
    {
        if (row == n)
        {
            vector<string> k;
            for (int i = 0; i < n; i++)
            {
                string s = "";
                for (int j = 0; j < n; j++)
                {
                    if (v[i][j] == 1)
                    {
                        s = s + 'Q';
                    }
                    else
                    {
                        s = s + '.';
                    }
                }
                k.push_back(s);
            }
            res.push_back(k);
            return;
        }

        for (int i = 0; i < n; i++)
        {
            if (isSafe(row, i, n, v))
            {
                v[row][i] = 1;
                findPos(row + 1, n, v);
                v[row][i] = 0;
            }
        }

        return;
    }

    vector<vector<string>> solveNQueens(int n)
    {

        vector<vector<int>> v(n, vector<int>(n, 0));

        findPos(0, n, v);
        return res;
    }
};
```

**Time Complexity: O(n!)**