<p>The <strong>n-queens</strong> puzzle is the problem of placing <code>n</code> queens on an <code>n x n</code> chessboard such that no two queens attack each other.</p>

<p>Given an integer <code>n</code>, return <em>the number of distinct solutions to the&nbsp;<strong>n-queens puzzle</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/queens.jpg" style="width: 600px; height: 268px;" />
<pre>
<strong>Input:</strong> n = 4
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are two distinct solutions to the 4-queens puzzle as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 9</code></li>
</ul>

class Solution {
public:
    int giveCount(vector<vector<bool>> avail, int n, int r){
        if(r == n - 1){
            int base = 0;
            for(int i=0; i<n; i++){
                if(avail[r][i]){
                    base++;
                }
            }

            cout<<base<<endl;
            return base;
        }

        int tot_count = 0;
        for(int i=0; i<n; i++){
            if(!(avail[r][i])){continue;}

            vector<vector<bool>> copy = avail;
            int c = i;
            for(int j=0; j<n; j++){
                copy[r][j] = 0;
                copy[j][c] = 0;
            }
            for(int j=(r + 1); j<n; j++){
                if((c + (j - r)) < n){copy[j][c + (j - r)] = 0;}
                if((c - (j - r)) >= 0){copy[j][c - (j - r)] = 0;}
            }

            tot_count += giveCount(copy, n, r + 1);
        }

        return tot_count;
    }

    int totalNQueens(int n) {
        vector<vector<bool>> avail(n, vector<bool> (n, 1));

        int count = giveCount(avail, n, 0);

        return count;
    }
};
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