<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<pair<int,pair<int,int>>> q;
        int m = mat.size();
        int n = mat[0].size();
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(mat[i][j]==0) q.push({0, {i,j}});
            }
        }
        vector<vector<int>> visited(m, vector<int>(n, 0));
        vector<vector<int>> dir{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        while(!q.empty()){
            auto node = q.front();
            int i = node.second.first;
            int j = node.second.second;
            int dist = node.first;
            q.pop();
            if(visited[i][j]!=0) continue;
            visited[i][j] = dist;
            for(int k=0;k<4;k++){
                int r = i + dir[k][0];
                int c = j + dir[k][1];
                if(r<0 || c<0 || r>=m || c>=n){
                    continue;
                }
                if(mat[r][c]==0 || visited[r][c]!=0) continue;
                q.push({dist+1, {r,c}});
            }
        }


        return visited;
    }
};
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