<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>>& grid) {
        int n = grid.size(),m = grid[0].size();
        vector<vector<bool>> vis(n,vector<bool>(m,false));
        queue<int> que;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j] == 0){
                    que.push(i*m+j);
                    vis[i][j] = true;
                }
            }
        }
        vector<vector<int>> dir {{-1,0},{1,0},{0,-1},{0,1}};
        while(!que.empty()){
            int size = que.size();
            while(size-- > 0){
                int idx = que.front();
                que.pop();
                int sr = idx/m;
                int sc = idx%m;
                for(auto &d : dir){
                    int r = sr + d[0];
                    int c = sc + d[1];
                    if(r >= 0 && c >= 0 && r < n && c < m && !vis[r][c]){
                        grid[r][c] = grid[sr][sc]+1;
                        vis[r][c] = true;
                        que.push(r*m+c);
                    }
                }
            }
        }
        return grid;
    }
};
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