<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();
        vector<vector<int>> ans(n, vector<int>(m, 1e9));
        for(int i = 0; i < n;i++){
            for(int j = 0; j < m;j++){
                if(mat[i][j] == 0)ans[i][j] = 0;
            }
        }
        for(int i = 0; i < n;i++){
            for(int j = 0; j < m;j++){
                if(mat[i][j] == 0)continue;
                vector<int> arr = {-1, 0, 1, 0, -1};
                for(int k =0 ;k < 4;k++){
                    int _x = i + arr[k];
                    int _y = j + arr[k + 1];
                    if(_x >=0 && _x < n && _y>= 0 && _y < m){
                        ans[i][j] = min(ans[i][j], ans[_x][_y] + 1);
                    }
                }
            }
        }
        for(int i = n-1; i >=0;i--){
            for(int j = m-1; j >=0;j--){
                if(mat[i][j] == 0)continue;
                vector<int> arr = {-1, 0, 1, 0, -1};
                for(int k =0 ;k < 4;k++){
                    int _x = i + arr[k];
                    int _y = j + arr[k + 1];
                    if(_x >=0 && _x < n && _y>= 0 && _y < m){
                        ans[i][j] = min(ans[i][j], ans[_x][_y] + 1);
                    }
                }
            }
        }
        return ans;
    }
};
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