<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:
void bfs(vector<vector<int>> mat,queue<pair<int,pair<int,int>>> zeros,vector<vector<int>>& matnew){
    vector<pair<int,int>> trav={{1,0},{-1,0},{0,1},{0,-1}};
       while(!zeros.empty()){
        int r=zeros.front().second.first;
        int l=zeros.front().first+1;
        int c=zeros.front().second.second;
        for(int i=0;i<4;i++){
            int newr=r+trav[i].first;
            int newc=c+trav[i].second;
            if(newr>-1 && newr<mat.size() && newc>-1 && newc<mat[0].size() ){
                  if(mat[newr][newc]!=0 && l<matnew[newr][newc]){
                    matnew[newr][newc]=l;
                    zeros.push(make_pair(l,make_pair(newr,newc)));
                  }
            }
        }
        zeros.pop();
       }
}
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
    vector<vector<int>> matnew(mat.size(),vector<int>(mat[0].size(),INT_MAX));
        queue<pair<int,pair<int,int>>> zeros;
        for(int i=0;i<mat.size();i++){
            for(int j=0;j<mat[0].size();j++){
                if(mat[i][j]==0) {zeros.push(make_pair(0,make_pair(i,j)));
                matnew[i][j]=0;
                }
            }
        }
       bfs(mat,zeros,matnew);
        return matnew;
    }
};
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