<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:
    bool check(vector<vector<int>>& mat, int i, int j, int m, int n, vector<vector<bool>>&visited){
        if (i >= 0 && i < m && j >= 0 && j < n){
            if (mat[i][j] == 1 && visited[i][j] == false){
                return true;
            }
        }
        return false;
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int m = mat.size();
        int n = mat[0].size();
        vector<vector<int>>ans(m, vector<int>(n, 0));
        vector<vector<bool>>visited(m, vector<bool>(n, false));
        queue<pair<int, pair<int, int>>>q;
        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                if (mat[i][j] == 0){
                    visited[i][j] = true;
                    q.push({0, {i, j}});
                }
            }
        }
        int dr[] = {0, 0, 1, -1};
        int dc[] = {1, -1, 0, 0};
        while (!q.empty()){
            int dis = q.front().first;
            int x = q.front().second.first;
            int y = q.front().second.second;
            q.pop();
            for (int i = 0; i < 4; i++){
                int newx = x + dr[i];
                int newy = y + dc[i];
                if (check(mat, newx, newy, m, n, visited)){
                    visited[newx][newy] = true;
                    ans[newx][newy] = dis+1;
                    q.push({dis+1, {newx, newy}});
                }
            }
        }
        return ans;
    }
};
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