<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:
    void bfs(vector<vector<int>> mat, vector<vector<int>> &vis,vector<vector<int>>&ans,queue<pair<pair<int,int>,int>>&q){
        int delrow[]={1,0,-1,0};
        int delcol[]={0,1,0,-1};
        while(!q.empty()){
            int size= q.size();
            for(int i=0;i<size;i++){
                int x = q.front().first.first;
                int y = q.front().first.second;
                int time = q.front().second;
                q.pop();
                vis[x][y]=1;
                ans[x][y]=time;
                for(int k=0;k<4;k++){
                    int nrow=x+delrow[k];
                    int ncol=y+delcol[k];
                    if(nrow>=0&&ncol>=0&&nrow<mat.size()&&ncol<mat[0].size()&&!vis[nrow][ncol]){
                        q.push({{nrow,ncol},time+1});
                        vis[nrow][ncol] = 1;
                    }
                }
            }

            
        }
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int n= mat.size();
        int m=mat[0].size();
        vector<vector<int>> vis(n,vector<int>(m,0));
        vector<vector<int>> ans(n,vector<int>(m,-1));
        int time=0;
        queue<pair<pair<int,int>,int>>q;
        
        
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mat[i][j]==0){ 
                    q.push({{i,j},0});
                    vis[i][j]=1;
                    // cout<<" "<<i<<" "<<j<<endl;
                }
            }
        }
        bfs(mat,vis,ans,q);
        return ans;
        
    }
};
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