<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();

        vector<vector<int>> result(m, vector<int>(n,0));
        vector<vector<int>> visited(m, vector<int>(n, 0));
        vector<pair<int,int>> zeroes;
        queue<pair<pair<int,int>,int>>q;
        int visit_val = 0;

        for(int i =0 ; i< m ; i++){
            for(int j =0 ; j< n ; j++){

                if(mat[i][j] == 0){
                    zeroes.push_back({i,j});
                    q.push({{i,j},0});
                }
            }
        }
                

        while(!q.empty()){

            pair<pair<int,int>,int> p = q.front();
            q.pop();
                        
            if(visited[p.first.first][p.first.second] == 0){
                                
                result[p.first.first][p.first.second] =p.second;

                if(p.first.first > 0){
                    q.push({{p.first.first-1,p.first.second},p.second+1});
                }
                if(p.first.first < m-1){
                    q.push({{p.first.first+1,p.first.second},p.second+1});
                }
                if(p.first.second > 0){
                    q.push({{p.first.first,p.first.second-1},p.second+1});
                }
                if(p.first.second < n-1){
                    q.push({{p.first.first,p.first.second+1},p.second+1});
                }
                visited[p.first.first][p.first.second] = 1;
            }
        }return result;
    }
};
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