<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(), e = 1e5;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if (mat[i][j] != 0) {
                    int top,left;
                    if (j == 0) left = e;
                    else left = mat[i][j-1];
                    if (i == 0) top = e;
                    else top = mat[i-1][j];
                    
                    mat[i][j] = min(top, left) + 1;
                }
            }
        }

        for(int i=n-1; i>=0; i--) {
            for(int j=m-1; j>=0; j--) {
                if (mat[i][j] != 0) {
                    int right,bottom;
                    if (j == m-1) right = e;
                    else right = mat[i][j+1];
                    if (i == n-1) bottom = e;
                    else bottom = mat[i+1][j];
                    
                    mat[i][j] = min(mat[i][j], min(bottom, right) + 1);
                }
            }
        }
        return mat;
    }
};
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