<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 {
    int m, n;
    bool valid(int r, int c){
        return r>=0 && r<m && c>=0 && c<n;
    }
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        m = mat.size(), n = mat[0].size();
        vector<vector<int>>res(m, vector<int>(n, INT_MAX));
        deque<vector<int>>q;
        for(int i = 0; i<m; i++){
            for(int j = 0; j<n; j++){
                if(mat[i][j]==0){
                    res[i][j] = 0;
                    q.push_back({0, i, j});
                }
            }
        }

        int x_c[] = {0, 0, 1, -1};
        int y_c[] = {1, -1, 0, 0};
        while(!q.empty()){
            const auto &v = q.front();
            int lvl = v[0], x = v[1], y = v[2];
            q.pop_front();
            
            for(int i = 0; i<=3; i++){
                int r = x_c[i] + x, c = y_c[i] + y;
                if(valid(r, c) && res[r][c]>lvl){
                    q.push_back({lvl+1, r, c});
                    res[r][c] = lvl+1;
                }
            }
        }
        return res;
    }
};
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