<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>> dist(m,vector<int>(n,INT_MAX));
        queue<pair<int,int>> q;
        for (int i=0;i<m;i++){
            for (int j=0;j<n;j++){
                if (mat[i][j]==0){
                    dist[i][j]=0;
                    q.push({i,j});
                }

            }
        }

        while(!q.empty()){
            auto [r,c]=q.front();
            q.pop();
            vector<pair<int,int>> directions={{1,0},{0,1},{-1,0},{0,-1}};
            for (auto [rd,cd]:directions){
                int nr=r+rd;
                int nc=c+cd;
                if (nr>=0 && nr<m && nc>=0 && nc<n){
                    if (dist[nr][nc]>dist[r][c]){
                        dist[nr][nc]=1+dist[r][c];
                        q.push({nr,nc});
                    }
                }
                
            }
        }

        return dist;
    }
};
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