<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();
        int m=mat[0].size();
        vector<vector<int>>dp(n,vector<int>(m,INT_MAX-2));
        //normal case so will go for top and left only
        for (int i=0;i<n;i++){
            for (int j=0;j<m;j++){
                if (mat[i][j]==0) dp[i][j]=0;
                else {
                    if (i>0) dp[i][j]= min (dp[i][j], dp[i-1][j]+1); //top
                    if (j>0) dp[i][j]=min (dp[i][j],dp[i][j-1]+1); //left
                }
            }
        }

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

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