<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:
    void bfs(queue<pair<pair<int , int> , int>>&q, vector<vector<int>>&grid , vector<vector<int>>&distance, int a, int b){
        while(!q.empty()){
            int i = q.front().first.first;
            int j = q.front().first.second;
            int k = q.front().second;
            q.pop();
            if(distance[i][j]!=-1) continue;
            distance[i][j] = k;
            if(i+1<a){
                q.push({{i+1 , j} , k+1});
            }
            if(i-1>=0){
                q.push({{i-1 , j} , k+1});
            }
            if(j+1<b){
                q.push({{i, j+1} , k+1});
            }
            if(j-1>=0){
                q.push({{i , j-1} , k+1});
            }
        }
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& grid){
        int a = grid.size();
        int b = grid[0].size();
        vector<vector<int>>distance(a , vector<int>(b ,-1));
        queue<pair<pair<int , int> , int>>q;
        for(int i =0 ; i<a ; i++){
            for(int j=0 ; j<b; j++){
                if(grid[i][j] ==0){
                    q.push({{i , j} , 0});    
                }
            }
        }
        bfs(q , grid , distance , a , b);
        return distance;
    }
};
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