<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>all distinct solutions to the <strong>n-queens puzzle</strong></em>. You may return the answer in <strong>any order</strong>.</p>

<p>Each solution contains a distinct board configuration of the n-queens&#39; placement, where <code>&#39;Q&#39;</code> and <code>&#39;.&#39;</code> both indicate a queen and an empty space, respectively.</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> [[&quot;.Q..&quot;,&quot;...Q&quot;,&quot;Q...&quot;,&quot;..Q.&quot;],[&quot;..Q.&quot;,&quot;Q...&quot;,&quot;...Q&quot;,&quot;.Q..&quot;]]
<strong>Explanation:</strong> There exist two distinct solutions to the 4-queens puzzle as shown above
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> [[&quot;Q&quot;]]
</pre>

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

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

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        vector<vector<string>> o;
        vector<string> c(n);
        vector<vector<bool>> v(n, vector<bool>(n, false));
        vector<bool> h(n, false);
        dfs(n, 0, -1000, v, c, o, h);
        return o;
    }
    void dfs(int n, int i, int p, vector<vector<bool>>& v, vector<string>& c, vector<vector<string>>& o,
    vector<bool>&h) {
        if(i == n) {
            o.push_back(c);
            return;
        }
        for(int x = 0; x < n; x++){
            if(!h[x] && !v[i][x]){
                string r = "";
                for(int y = 0; y < n; y++){
                    r += y == x ? "Q" : ".";
                }
                c[i] = r; h[x] = true;
                stack<pair<int, int>> s;
                visit(n, i, x, v, s);
                dfs(n, i + 1, x, v, c, o, h);
                while(!s.empty()){
                    v[s.top().first][s.top().second] = false;
                    s.pop();
                }
                h[x] = false;
            }
        }
    }

    void visit(int n, int i, int k, vector<vector<bool>>& v, stack<pair<int, int>>& s) {
        int d = i; int l = k; int r = k;
        while(d < n && l > -1) {
            if(!v[d][l]) { s.push({d, l}); }
            v[d++][l--] = true;
        }
        d = i;
        while(d < n && r < n) {
            if(!v[d][r]) { s.push({d, r}); }
            v[d++][r++] = true;
        }
    }
};
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