<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:
map<pair<int,int>,int> multibfs(vector<pair<int,int>> start,vector<vector<int>>& grid)
    {
        map<pair<int,int>,int> dist;
        queue<pair<int,int>> q;
        int i,j,n,m;
        n = grid.size();
        m = grid[0].size();
         for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(grid[i][j]==1)
                {
                   dist[{i,j}] = 1e9;
                }
            }
        }
        for(i=0;i<start.size();i++)
        {
            q.push(start[i]);
            dist[start[i]]=0;
        }
        while(!q.empty())
        {
            pair<int,int> p = q.front();
            q.pop();
            int x = p.first;
            int y = p.second;
            if(x<n-1)
            {
                if(grid[x+1][y]==1 && dist[{x+1,y}]>dist[p]+1)
                {
                    q.push({x+1,y});
                    dist[{x+1,y}] = dist[p]+1;
                }
            }
            if(x>0)
            {
                if(grid[x-1][y]==1 &&dist[{x-1,y}]>dist[p]+1)
                {
                    q.push({x-1,y});
                    dist[{x-1,y}] = dist[p]+1;
                }
            }if(y<m-1)
            {
                if(grid[x][y+1]==1 &&dist[{x,y+1}]>dist[p]+1)
                {
                    q.push({x,y+1});
                    dist[{x,y+1}] = dist[p]+1;
                }
            }if(y>0)
            {
                if(grid[x][y-1]==1 && dist[{x,y-1}]>dist[p]+1)
                {
                    q.push({x,y-1});
                    dist[{x,y-1}] = dist[p]+1;
                }
            }
        }
        return dist;
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& grid) {
        int i,j,n,m;
        n = grid.size();
        m = grid[0].size();
        vector<pair<int,int>> start;
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(grid[i][j]==0)
                {
                    start.push_back({i,j});
                }
            }
        }
        map<pair<int,int>,int> dist = multibfs(start,grid);
        vector<vector<int>> ans(n,vector<int>(m));
        for(auto it: dist)
        {
            int x = it.first.first;
            int y = it.first.second;
            ans[x][y] = it.second;
        }
       
        return ans;
    }
    
};
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