<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:
    typedef pair<int, int> pi;
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int n = mat.size();
        int m = mat[0].size();
        vector<vector<int>> ans(n, vector<int>(m, 1e8));
        vector<vector<int>> vis(n, vector<int> (m, 0));
        queue<pair<int, pi>>q;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(mat[i][j] == 0){
                    q.push({0, {i, j}});
                    vis[i][j] = 1;
                }
            }
        }
        while(!q.empty()){
            int dis = q.front().first;
            int i = q.front().second.first;
            int j = q.front().second.second;
            vis[i][j] = 1;
            q.pop();
            if(mat[i][j] == 0) dis = 0;
            ans[i][j] = min(ans[i][j], dis);
            int a[] = {0, 0, -1, 1};
            int b[] = {1, -1, 0, 0};
            for(int k=0; k<4; k++){
                int x = i+a[k];
                int y = j+b[k];
                if(x >=0 && x<n && y>=0 && y<m && vis[x][y] != 1){
                    q.push({dis+1, {x, y}});
                }
            }
        }
        return ans;
    }
};
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