<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>> visited;
    void bfsmulti(vector<vector<int>>mat,vector<vector<int>>& output,vector<pair<int,int>> source)
    {
    queue <pair<pair<int,int>,int>> q;    
    for(auto[x,y]:source)
    {
        q.push({{x,y},0});
        visited[x][y]=1;
    }
    while(!q.empty())
    {
        auto[x,y]=q.front().first;
        int level=q.front().second;
        vector<pair<int,int>> neighbors={{x+1,y},{x,y-1},{x,y+1},{x-1,y}};
        q.pop();
        for(auto[i,j]:neighbors)
        {
            if(i>=0 && j>=0 && i<mat.size() && j<mat[0].size() && visited[i][j]==0)
            {
            visited[i][j]=1;
            output[i][j]=level+1;
            q.push({{i,j},level+1});
            }
        }
    }

    }

    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
    vector<pair<int,int>> source;
    for(int i=0;i<mat.size();i++)
    {
        for(int j=0;j<mat[0].size();j++)
        {
            if(mat[i][j]==0)
            {
                source.push_back({i,j});
            }
        }
    }
    vector<vector<int>> output(mat.size(),vector<int>(mat[0].size(),0));
    visited=vector<vector<int>>(mat.size(),vector<int>(mat[0].size(),0));
    bfsmulti(mat,output,source);
    return(output);
    }
};
58803