<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:
    int dr[4]={1,0,-1,0};
    int dc[4]={0,1,0,-1};
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int m = mat.size();
        int n = mat[0].size();
        map<pair<int,int>,int>vis;
        vector<vector<int>>ans(m,vector<int>(n,0));
        queue<pair<pair<int,int>,int>>q;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(mat[i][j]==0){
                    q.push({{i,j},0});
                    vis[{i,j}]=1;
                }
            }
        }
        while(!q.empty()){
            auto k = q.front();
            q.pop();
            int x = k.first.first;
            int y = k.first.second;
            int dis = k.second;
            for(int i=0;i<4;i++){
                int newX = x+dr[i];
                int newY = y+dc[i];
                if(newX>=0 and newX<m and newY>=0 and newY<n and !vis[{newX,newY}]){
                    q.push({{newX,newY},dis+1});
                    vis[{newX,newY}]=1;
                    ans[newX][newY]=dis+1;
                }
            }
        }
        return ans;
    }
};
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