<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>> dir = {{1,0},{0,1},{-1,0},{0,-1}} ;
    void bfs(queue<pair<int,int>>& q ,set<pair<int,int>>& visited,vector<vector<int>>& ans){
        int dist = 0 ;
        while(q.empty()==0){
            auto p = q.front() ;
            q.pop() ;
            if(p.first == -1 && p.second==-1){
                dist++ ;
                if(q.empty() == 0) q.push({-1,-1}) ;
                continue ;
            }
            //if((visited.find(p)!=visited.end())) continue ;
            visited.insert(p) ;
            if(ans[p.first][p.second] == -1)
            ans[p.first][p.second] = dist ;
            //else continue ;
            for(int i = 0 ; i < 4 ; i++){
                int r = p.first + dir[i][0] ;
                int c = p.second + dir[i][1] ;
                if(r<0 || c<0 || r==ans.size() || c==ans[0].size()) continue ;
                if(visited.find({r,c})==visited.end()) q.push({r,c}) ;
            }
        }
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int n = mat.size() , m = mat[0].size() ;
        vector<vector<int>> ans (n , vector<int> (m,-1)) ;
        queue<pair<int,int>> q ;
        set<pair<int,int>> s ;
        for(int i = 0 ; i< n ; i++){
            for(int j = 0. ;j < m ; j ++){
                if(mat[i][j]==0){ 
                    ans[i][j] = 0 ;
                    q.push({i,j});
                    s.insert({i,j});
                }
            }
        }
        q.push({-1,-1}) ;
        bfs(q,s,ans) ;
        return ans ;
    }
};
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