<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(); 
        int m = mat[0].size();
        set<pair<int, int>> q;
        vector<int> foo (m, -1);
        vector<vector<int>> ret (n, foo);
        
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(mat[i][j] == 0) {
                    q.insert(make_pair(i, j));
                    ret[i][j] = 0;
                }
            }
        }
        int layer = 0;
        while(!q.empty()) {
            set<pair<int, int>> next;
            for(pair<int, int> p: q) {
                int x = p.first;
                int y = p.second;
                if(x > 0 && ret[x-1][y] == -1) {
                    next.insert(make_pair(x-1, y));
                    ret[x-1][y] = layer + 1;
                }
                if(x < n-1 && ret[x+1][y] == -1) {
                    next.insert(make_pair(x+1, y));
                    ret[x+1][y] = layer + 1;
                }
                if(y > 0 && ret[x][y-1] == -1) {
                    next.insert(make_pair(x, y-1));
                    ret[x][y-1] = layer + 1;
                }
                if(y < m-1 && ret[x][y+1] == -1) {
                    next.insert(make_pair(x, y+1));
                    ret[x][y+1] = layer + 1;
                }
            }
            q = next;
            layer++;
        }
        return ret;
    }
};
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