<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>> updateMatrix(vector<vector<int>>& mat) {
        int m=mat.size(),n=mat[0].size();
        vector<vector<int>>vis(m,vector<int>(n,-1));
        queue<pair<int,int>>q;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(mat[i][j]==0)
                {
                    vis[i][j]=0;
                    q.push({i,j});
                }
            }    
        }
        int num=0;
        while(!q.empty())
        {
            queue<pair<int,int>>q1;
            q1=q;
            if(!q.empty())num++;
            while(!q.empty())q.pop();
            while(!q1.empty())
            {
                int i=q1.front().first,j=q1.front().second;
                q1.pop();
                if(i>0 && mat[i-1][j]==1 && vis[i-1][j]==-1)
                {
                    vis[i-1][j]=num;
                    q.push({i-1,j});
                }    
                if(j>0 && mat[i][j-1]==1 && vis[i][j-1]==-1)
                {
                    vis[i][j-1]=num;
                    q.push({i,j-1});
                } 
                if(i<m-1 && mat[i+1][j]==1 && vis[i+1][j]==-1)
                {
                    vis[i+1][j]=num;
                    q.push({i+1,j});
                }
                if(j<n-1 && mat[i][j+1]==1 && vis[i][j+1]==-1)
                {
                    vis[i][j+1]=num;
                    q.push({i,j+1});
                }     
            }
        }
        return vis;
    }
};
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