<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) {
        size_t m = mat.size();
        size_t n = mat[0].size();
        
        queue<vector<int>> currLevel;
        
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(mat[i][j]==0) currLevel.push({i,j});
                mat[i][j]*=-1;
            }
        }
        
        int currDist=1;
        while(!currLevel.empty()){
            size_t s = currLevel.size();
            for(size_t a=0; a<s; a++){
                vector<int> currPos = currLevel.front();
                currLevel.pop();
                for(int i=-1; i<2; i++){
                    for(int j=-1; j<2; j++){
                        if(i*j!=0) continue;
                        int x = currPos[0]+i;
                        int y = currPos[1]+j;
                        if(x<0 || x>m-1) continue;
                        if(y<0 || y>n-1) continue;
                        if(mat[x][y]>=0) continue; //been here before
                        mat[x][y]=currDist;
                        currLevel.push({x,y});
                    }
                }
            }
            currDist++;  
        }
        
        return mat;
        
    }
};
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