<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(vector<vector<int>>& mat,vector<vector<int>>&ans,queue<pair<int,int>>&q){
    int level=0;
   while(q.size()>0){
        auto curr=q.front();
        int r=curr.first;
        int c=curr.second;
        q.pop();
        if(r==200 && c==200){
            if(q.size()!=0){
                q.push({200,200});
                level++;
            }
            else{
               return;
            }
        }
        for(int i=0;i<4;i++){
            int nr=r+dir[i][0];
            int nc=c+dir[i][1];
            if(nr<0 || nc<0 ||nr>=mat.size() ||nc>=mat[0].size()||ans[nr][nc]!=INT_MAX) continue;
            ans[nr][nc]=min(ans[nr][nc],level+1);
            q.push({nr,nc});
        }
    }
}
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int r=mat.size();
        int c=mat[0].size();
        vector<vector<int>>ans(r,vector<int>(c,INT_MAX));
        queue<pair<int,int>>q;
        for(int i=0;i<r;i++){
            for(int j=0;j<c;j++){
               if(mat[i][j]==0){
                q.push({i,j});
                ans[i][j]=0;
               }
            }
        }
        q.push({200,200});
        bfs(mat,ans,q);
        return ans;
    }
};
34469