<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>> bfs(queue<pair<pair<int,int>, int>> q, vector<vector<int>>& adj, vector<vector<int>>& vis){
        vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        int n = adj.size();
        int m = adj[0].size();
        vector<vector<int>> res(n, vector<int>(m, 0));


        while(!q.empty()){
            int r = q.front().first.first;
            int c = q.front().first.second;
            int t = q.front().second;
            q.pop();

            for(auto d : directions) {
                int ni = r + d.first;
                int nj = c + d.second;

                if(ni >= 0 && ni < n && nj >= 0 && nj < m && adj[ni][nj] == 1 && !vis[ni][nj]){
                    vis[ni][nj] = 1;
                    res[ni][nj] = t + 1;
                    q.push({{ni, nj}, t + 1});    
                }
            }
        }
        return res;
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        vector<vector<int>> vis(mat.size(), vector<int>(mat[0].size(),0));
        queue<pair<pair<int,int>, int>> q;
        for(int i = 0; i < mat.size(); i++){
            for(int j = 0; j < mat[0].size(); j++){
                if(mat[i][j] == 0){
                    vis[i][j] = 1;
                    q.push({{i, j}, 0});
                }
            }
        }
        vector<vector<int>> arr = bfs(q, mat, vis);
        return arr;
    }
};
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