<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 n= mat.size(), m = mat[0]. size();
        vector< vector<int> >ans(n,vector<int> (m,INT_MAX));

        queue<vector<int>>q;

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

      int rows[]= {-1 ,0 ,1, 0};
      int cols[]= {0, -1, 0 ,1};

      while(!q.empty())
      {
        int row = q.front()[0];
        int col = q.front()[1];
        int dis = q.front()[2];
        q.pop();

        for(int i=0; i<=3; i++)
        {
            int r = row + rows[i];
            int c = col + cols[i];

            if(r>=0 && r<n && c>=0 && c<m)
            {
                if(mat[r][c] == 1 && dis+1<ans[r][c])
                {
                    q.push({r , c, dis+1});
                    ans[r][c]=dis+1;
                }
            }
        }
      }
      return ans;
        
    }
};
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