<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();
        int n = mat[0].size();
        map<pair<int,int>, bool> visited;
        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});
                    visited[{i,j}] = true;
                } else {
                    mat[i][j] = -1; 
                }
            }
        }
        while(!q.empty()){
            int size = q.size();
            
            for( int k = 0 ; k< size;k++){
                auto top = q.front();
                q.pop();
                int i = top.first.first;
                int j = top.first.second;
                int dist = top.second;

                int dx[] = {-1,0,1,0};
                int dy[] = {0,1,0,-1};
                for( int it = 0 ; it<4;it++){
                    int newX = i + dx[it];
                    int newY = j + dy[it];
                    if( newX >= 0 && newY >= 0 && newX < m && newY < n && 
                        mat[newX][newY] != 0 && !visited[{newX,newY}]){
                        q.push({{newX,newY},dist+1});
                        visited[{newX,newY}] = true;
                        mat[newX][newY] = dist + 1;

                    }
                }
            }
        }
        return mat;
    }
};
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