<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>> res = mat;
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        queue<pair<int,int>> q;

        for(int i = 0; i < m ; i++){
            for(int j = 0; j < n; j++){
                if(!mat[i][j]){
                    q.push({i,j});
                    visited[i][j] = true;
                }
                
            }
        }
        int ans = 0;

        while(!q.empty()){
            int size = q.size();
            while(size--){
                auto front = q.front();
                q.pop();

                res[front.first][front.second] = ans;

                vector<pair<int,int>> dir = {{0,1}, {1,0}, {0,-1}, {-1,0}};
                for(int i= 0; i < 4; i++){
                    int new_i = dir[i].first + front.first;
                    int new_j = dir[i].second + front.second;

                    if(new_i >= 0 && new_j >= 0 && new_i < m && new_j < n && mat[new_i][new_j] && !visited[new_i][new_j]){
                        q.push({new_i, new_j});
                        visited[new_i][new_j] = true;
                    }
                }
            }
            ans++;
        }

        return res;
    }
};
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