<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 {
    vector<pair<int,int>>dirs = { {1,0},{0,1},{-1,0},{0,-1} };
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        
        list<pair<int,pair<int,int>>>myList;   // time and coords 
        int rows = mat.size();
        int cols = mat[0].size();
        
        for( int i=0; i<rows; i++ ){
            for( int j = 0; j<cols; j++ ){
                if( mat[i][j] == 1 ){
                    mat[ i ][ j ] = -1;
                } else {
                    myList.push_back( { 0, { i, j } } );
                }
            }
        }
        
        
        while( myList.size() ){
            
            pair<int,pair<int,int>>topElem = myList.front();
            myList.pop_front();
            
            for( int k=0; k<dirs.size(); k++ ){
                
                int newi = topElem.second.first + dirs[k].first;
                int newj = topElem.second.second + dirs[k].second;
                
                if( newi >= 0 && newj >=0 && newi < rows && newj < cols && mat[newi][newj] == -1 ){
                    myList.push_back( { topElem.first + 1, { newi, newj } } );
                    mat[newi][newj] = max( mat[newi][newj], topElem.first + 1 );
                }
            }
        
        }
        

        return mat;
        
        
        
        
        
    }
};
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