<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:
   int ret(vector<vector<int>>& mat, int i, int j, vector<vector<int>>& dp) {
      if (i < 0 || i>=mat.size() || j<0 || j>=mat[0].size() || mat[i][j] == 2)
        return 1e5;
      if (mat[i][j] == 0) return 0;
      if (dp[i][j] != 0) return dp[i][j];
      mat[i][j] = 2; 
      return dp[i][j] = 1+ min(ret(mat,i,j-1, dp),
      min(ret(mat,i-1,j, dp),
      min(ret(mat,i,j+1, dp),
      ret(mat,i+1,j, dp))));
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        vector<vector<int>>dp(mat.size(), vector<int>(mat[0].size(), 0));
        for(int i=0; i<mat.size(); i++) {
          for(int j=0; j<mat[0].size(); j++) {
            if (mat[i][j] && dp[i][j] == 0) {
              ret(mat, i,j, dp);
            }
          }
        }
        
        for(int i=0; i<mat.size(); i++) {
            for(int j=0; j<mat[0].size(); j++) {
                int left = (j > 0) ? dp[i][j-1] + 1 : INT_MAX;
                int top = (i > 0) ? dp[i-1][j] + 1 : INT_MAX;
                int right = (j < dp[0].size() - 1) ? dp[i][j+1] + 1 : INT_MAX;
                int bottom = (i < dp.size() - 1) ? dp[i+1][j] + 1 : INT_MAX;
                dp[i][j] = min(dp[i][j], min(top, min(left, min(right, bottom))));
            }
        }
        for(int i=mat.size() - 1; i>=0; i--) {
            for(int j=mat[0].size() - 1; j>=0; j--) {
                int left = (j > 0) ? dp[i][j-1] + 1 : INT_MAX;
                int top = (i > 0) ? dp[i-1][j] + 1 : INT_MAX;
                int right = (j < dp[0].size() - 1) ? dp[i][j+1] + 1 : INT_MAX;
                int bottom = (i < dp.size() - 1) ? dp[i+1][j] + 1 : INT_MAX;
                dp[i][j] = min(dp[i][j], min(top, min(left, min(right, bottom))));
            }
        }
        return dp;
    }
};
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