<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>>up(m,vector<int>(n,m*n));
        vector<vector<int>>down(m,vector<int>(n,m*n));
        vector<vector<int>>left(m,vector<int>(n,m*n));
        vector<vector<int>>right(m,vector<int>(n,m*n));
        vector<vector<int>>ans(m,vector<int>(n,m*n));

        for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++) {
                if(mat[i][j]==0) {up[i][j]=0;continue;}
                if(i!=0)
                up[i][j]=min(up[i][j],up[i-1][j]+1);
                if(j!=0)
                up[i][j]=min(up[i][j],up[i][j-1]+1);
            }
        }

        for(int i=m-1;i>=0;i--) {
            for(int j=n-1;j>=0;j--) {
                if(mat[i][j]==0) {up[i][j]=0;continue;}
                if(i!=m-1)
                up[i][j]=min(up[i][j],up[i+1][j]+1);
                if(j!=n-1)
                up[i][j]=min(up[i][j],up[i][j+1]+1);
            }
        }

        // for(int j=0;j<n;j++) {
        //     for(int i=0;i<m;i++) {
        //         if(mat[i][j]==0) {left[i][j]=0;continue;}
        //         if(j!=0)
        //         left[i][j]=min(left[i][j],left[i][j-1]+1);
        //         if(i!=0)
        //         left[i][j]=min(left[i][j],left[i-1][j]+1);
        //     }
        // }
        
        // for(int j=n-1;j>=0;j--) {
        //     for(int i=0;i<m;i++) {
        //         if(mat[i][j]==0) {right[i][j]=0;continue;}
        //         if(j!=n-1)
        //         right[i][j]=min(right[i][j],right[i][j+1]+1);
        //         if(i!=0)
        //         right[i][j]=min(right[i][j],right[i-1][j]+1);
        //     }
        // }

        // for(int i=0;i<m;i++) {
        //     for(int j=0;j<n;j++) cout<<up[i][j]<<" ";
        //     cout<<endl;
        // }

        for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++) {
                ans[i][j]=min(up[i][j],min(down[i][j],min(left[i][j],right[i][j])));
            }
        }

        return ans;


    }
};
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