<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) {
    int m = mat.size();
    int n = mat[0].size();
    vector<vector<int>> res(m, vector<int>(n, 0));
    vector<vector<bool>> visited(m, vector<bool>(n, false));
    int r[] = {1, -1, 0, 0};
    int c[] = {0, 0, -1, 1};
    queue<pair<int, pair<int, int>>>q;
    for(int i = 0; i< m; i++){
        for(int j = 0; j < n; j++){
            if(mat[i][j] ==0){
                visited[i][j] = true;
                q.push({0, {i, j}});
            }
        }
    }
    while(!q.empty()){
        auto temp = q.front();
        int x = temp.second.first;
        int y = temp.second.second;
        int dist = temp.first;
        res[x][y] = dist;
        q.pop();
        for(int k=0; k<4; k++){
            int i = x + r[k];
            int j = y + c[k];
            if(i>=0 && j >=0 && i < m && j < n && visited[i][j]!=true){
                visited[i][j]=true;
                q.push({dist+1, {i, j}});
            }
        }
    }
    return res;    
}
};
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