<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>> ans(m,vector<int>(n));
        vector<vector<bool>> visited(m,vector<bool>(n));
       
       queue<vector<int>> q;

       for(int i=0;i<m;i++)
          for(int j=0;j<n;j++)
          {
             if(mat[i][j]==0){
                ans[i][j]=0;
                q.push({0,i,j});
                visited[i][j]=true;
             }

          }


        int ai[4]={-1,0,1,0};
        int aj[4]={0,-1,0,1};

         
      while(!q.empty()){
        
        int dist=q.front()[0];
        int i=q.front()[1];
        int j=q.front()[2];
        q.pop();

        for(int k=0;k<4;k++){
           int new_i=ai[k]+i;
           int new_j=aj[k]+j;
            if(new_i>=0&&new_j>=0&&new_i<m&&new_j<n&&visited[new_i][new_j]==false){
             q.push({dist+1,new_i,new_j});
               visited[new_i][new_j]=true;
               ans[new_i][new_j]=dist+1;
            }

        }
      }
      


      return ans;
   
    
             
    }
};
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