<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>> distance;
    int value(int row, int col) {
        if (row < 0 || col < 0 || row >= distance.size() ||
            col >= distance[0].size()) {
            return INT_MAX;
        }
        return distance[row][col];
    }
    void solve(vector<vector<int>>& mat, int row, int col) {
        if (row < 0 || col < 0 || row >= distance.size() ||
            col >= distance[0].size()) {
            return;
        }
        if (mat[row][col] == 0)
            distance[row][col] = 0;
        else {
            int dis = min(value(row,col-1),min(value(row,col+1),min(value(row+1,col),value(row-1,col))));
            distance[row][col] = dis == INT_MAX ? INT_MAX : dis + 1;
        }
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        distance.resize(mat.size(),vector<int>(mat[0].size(),INT_MAX));
        for(int row = 0 ; row < mat.size() ; row++){
            for(int col = 0 ; col < mat[0].size() ; col++){
                solve(mat,row,col);
            }
        }
        for(int row = mat.size() - 1; row >= 0 ; row--){
            for(int col = mat[0].size() - 1; col >= 0 ; col--){
                solve(mat,row,col);
            }
        }
        return distance;
    }
};
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