<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 { // 48 ms, faster than 99.64%
public:
    vector<vector<int>> updateMatrix(vector<vector<int>> &mat) {
        int m = mat.size(), n = mat[0].size(), INF = m + n; // The distance of cells is up to (M+N)
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (mat[r][c] == 0) continue;
                int top = INF, left = INF;
                if (r - 1 >= 0) top = mat[r - 1][c];
                if (c - 1 >= 0) left = mat[r][c - 1];
                mat[r][c] = min(top, left) + 1;
            }
        }
        for (int r = m - 1; r >= 0; r--) {
            for (int c = n - 1; c >= 0; c--) {
                if (mat[r][c] == 0) continue;
                int bottom = INF, right = INF;
                if (r + 1 < m) bottom = mat[r + 1][c];
                if (c + 1 < n) right = mat[r][c + 1];
                mat[r][c] = min(mat[r][c], min(bottom, right) + 1);
            }
        }
        return mat;
    }
};
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