<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>>& v) {
        int n=v.size(),m=v[0].size();
        vector<vector<int>>visited(n,vector<int>(m,0));
        queue<vector<int>>q;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(!v[i][j]){
                    q.push({0,i,j});
                    visited[i][j]++;
                }
            }
        }
        int drow[]={0,0,1,-1},dcol[]={1,-1,0,0};
        while(q.size()){
            int dis=q.front()[0];
            int row=q.front()[1];
            int col=q.front()[2];
            q.pop();
            for(int i=0;i<4;i++){
                int nrow=row+drow[i],ncol=col+dcol[i];
                if(nrow>=0&&ncol>=0&&nrow<n&&ncol<m&&!visited[nrow][ncol]){
                    visited[nrow][ncol]++;
                    v[nrow][ncol]=dis+1;
                    q.push({dis+1,nrow,ncol});
                }
            }
        }
        return v;
    }
};
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