<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> > dist(n, vector<int>(m, INT_MAX));
        priority_queue< pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>> >pq;

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

        while(!pq.empty()){
            auto [current, config] = pq.top();
            pq.pop();
            auto [row, col] = config;

            if( dist[row][col] < current){
                continue;
            }

            vector<int> arrx = {1, -1, 0, 0};
            vector<int> arry = {0, 0, 1, -1};

            for(int i=0; i<4; i++){
                int newrow = row + arrx[i];
                int newcol = col + arry[i];

                if(newrow>=0 && newcol>=0 && newrow<n && newcol<m && mat[newrow][newcol] == 1 && dist[newrow][newcol] > current+1){
                    dist[newrow][newcol] = current + 1;
                    pq.push({dist[newrow][newcol], {newrow, newcol}});
                }
            }
            
        }

        return dist;
    }
};
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