<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<pair<int,int>,int>>q;
        int rows=mat.size();
        int cols=mat[0].size();
        vector<vector<int>>visited(rows,vector<int>(cols,0));
        
        vector<vector<int>>distance(rows,vector<int>(cols,0));
        for(int i=0;i<rows;i++){
            for(int j=0;j<cols;j++){
                if(mat[i][j]==0){
                    q.push({{i,j},0});
                    visited[i][j]=1;
                }
            }
        }
        while(!q.empty()){
            int x=q.front().first.first;
            int y=q.front().first.second;
            int steps=q.front().second;
            vector<pair<int,int>>v{{x+1,y},{x,y+1},{x-1,y},{x,y-1}};
            for(auto p:v){
                int l=p.first;int m=p.second;
                if(l>=0&&l<rows&&m>=0&&m<cols){
                    if(mat[l][m]==1&&visited[l][m]!=1){
                        q.push({{l,m},steps+1});
                        visited[l][m]=1;
                    }
                }
            }
            
            distance[x][y]=steps;
            q.pop();
        }
        return distance;
    }
};
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