<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) {
        queue<pair<int,pair<int,int>>> q;
        map<pair<int,int>,int> vis;
        int n = mat.size();
        int m = mat[0].size();
        for(int i=0;i<mat.size();i++)
        {
            for(int j=0;j<mat[i].size();j++)
            {
                if(mat[i][j]==0)
                {
                    q.push({0,{i,j}});
                    vis[{i,j}]=0;
                }
                else
                vis[{i,j}]=INT_MAX;
            }
        }
        
        while(!q.empty())
        {
            int x = q.front().first;
            int y = q.front().second.first;
            int z = q.front().second.second;
            q.pop();
            if(y-1>=0 && vis[{y-1,z}]>1+x)
            {
                vis[{y-1,z}]=x+1;
                q.push({x+1,{y-1,z}});
            }
            if(z-1>=0 && vis[{y,z-1}]>1+x)
            {
                vis[{y,z-1}]=x+1;
                q.push({x+1,{y,z-1}});
            }
            if(y+1<n && vis[{y+1,z}]>1+x)
            {
                vis[{y+1,z}]=1+x;
                q.push({1+x,{y+1,z}});
            }
            if(z+1<m && vis[{y,z+1}]>1+x)
            {
                vis[{y,z+1}]=1+x;
                q.push({1+x,{y,z+1}});
            }
        }
        vector<vector<int>> ans(n,vector<int>(m));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                ans[i][j]=vis[{i,j}];
            }
        }
        return ans;
    }
};
53781