<p>Given an <code>m x n</code> binary matrix <code>mat</code>, return <em>the distance of the nearest </em><code>0</code><em> for each cell</em>.</p>

<p>The distance between two adjacent cells is <code>1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-1-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[0,0,0]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[0,0,0]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-2-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[1,1,1]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[1,2,1]]
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one <code>0</code> in <code>mat</code>.</li>
</ul>

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        queue <tuple<int,int,int>> q;
        vector<vector<int>> v(mat.size(),vector<int>(mat[0].size(),INT_MAX));
        for(int i = 0; i < mat.size(); i++){
            for(int j = 0; j < mat[i].size(); j++){
                if(mat[i][j] == 0) {
                    q.push({i,j,0});
                    v[i][j] = 0;
                }
            }
        }
        vector<vector<bool>> visited(mat.size(),vector<bool>(mat[0].size(),false));
        while(!q.empty()){
            auto it = q.front();
            q.pop();
            int r = get<0>(it);
            int c = get<1>(it);
            int d = get<2>(it);
            visited[r][c] = true;
            v[r][c] = min(d,v[r][c]);
            if(r > 0 && !visited[r-1][c]) q.push({r-1,c,d+1});
            if(c > 0 && !visited[r][c-1]) q.push({r,c-1,d+1});
            if(r < mat.size()-1 && !visited[r+1][c]) q.push({r+1,c,d+1});
            if(c < mat[r].size()-1 && !visited[r][c+1]) q.push({r,c+1,d+1});
        }
        return v;
    }
};
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