<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:
queue<pair<int,int>>q;   vector<vector<int>>dp;set<pair<int,int>>st;
vector<vector<int>>dir={{0,1},{0,-1},{1,0},{-1,0}};
int f(queue<pair<int,int>>&q,vector<vector<int>>&v){
int t=0;
while(!q.empty()){
int s=q.size();
while(s--){
auto  el=q.front();
int r=el.first;
int c=el.second;
dp[r][c]=t;//st.insert({r,c});
q.pop();
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>=v.size()||nc>=v[0].size()){continue;}
if(v[nr][nc]==0){continue;}
if(st.count({nr,nc})!=0){continue;}
st.insert({nr,nc});
q.push({nr,nc});
}

}

t++;
}


return t;
}
    vector<vector<int>> updateMatrix(vector<vector<int>>& v) {
        dp.resize(v.size(),vector<int>(v[0].size(),-1));
for(int i=0;i<v.size();i++){
for(int j=0;j<v[0].size();j++){
if(v[i][j]==0){
    dp[i][j]=0;q.push({i,j});
}

}

}  

f(q,v);
return dp;

    }
};
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