<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();
        queue<pair<pair<int,int>,int>>q;
                vector<vector<int>>vis(m,vector<int>(n));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(mat[i][j]==0){
                 q.push({{i,j},0});
                 vis[i][j]=1;
                }
            }
        }


        while(!q.empty()){
            int x=q.front().first.first;
            int y=q.front().first.second;
            int p=q.front().second;
            q.pop();
            vector<int>v1={0,1,0,-1};
            vector<int>v2={1,0,-1,0};
            for(int i=0;i<4;i++){
                int xx=x+v1[i];
                int yy=y+v2[i];
                if(xx>=0 && xx<m && yy>=0 && yy<n){
                    if(vis[xx][yy]==0){
                        mat[xx][yy]=p+1;
                        q.push({{xx,yy},p+1});
                        vis[xx][yy]=1;
                    }
                }
            }
        }
        return mat;

    }
};
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