<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 m = mat.size();
        int n = mat[0].size();
        vector<vector<int>> dp;
        for (vector<int>& row:mat){
            dp.emplace_back(row.begin(), row.end());
        }
        for (int row = 0; row<m; row++){
            for(int col = 0; col < n; col++){
                if(dp[row][col] == 0){
                    continue;
                }
                int minNeighbor = m*n;
                if(row > 0){
                    minNeighbor = min(minNeighbor, dp[row-1][col]);
                }
                if(col > 0){
                    minNeighbor = min(minNeighbor, dp[row][col-1]);
                }
                dp[row][col] = minNeighbor +1;
            }
        }
        for(int row = m-1; row >= 0; row--){
            for(int col = n-1; col >= 0; col--){
                if (dp[row][col] == 0){
                    continue;
                }
                int minNeighbor = m*n;
                if(row < m-1){
                    minNeighbor= min(minNeighbor, dp[row+1][col]);
                }
                if(col < n-1){
                    minNeighbor = min(minNeighbor, dp[row][col+1]);
                }
                dp[row][col] = min(dp[row][col], minNeighbor+1);
            }
        }
        return dp;
    }
};
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