<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 {
private:
    void bfs(vector<vector<int>>& mat, vector<vector<int>>& visited, vector<vector<int>>& dist){

        int n = mat.size();
        int m = mat[0].size();

        queue<pair<pair<int,int>,int>> q;

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mat[i][j] == 0){
                    visited[i][j] = 1;
                    q.push({{i,j},0});
                }
            }
        }

        while(!q.empty()){

            int row = q.front().first.first;
            int col = q.front().first.second;
            int steps = q.front().second;

            q.pop();

            int rowdir[] = {-1,1,0,0};
            int coldir[] = {0,0,-1,1};

            for(int i=0;i<4;i++){
                int nrow = row + rowdir[i];
                int ncol = col + coldir[i];

                if(nrow>=0 && nrow<n && ncol>=0 && ncol<m && mat[nrow][ncol]==1 && !visited[nrow][ncol]){

                    visited[nrow][ncol] = 1;
                    dist[nrow][ncol] = steps+1;

                    q.push({{nrow,ncol},steps+1});
                }
            }
        }
        
    }
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int n = mat.size();
        int m = mat[0].size();

        vector<vector<int>> visited(n, vector<int>(m,0));
        vector<vector<int>> dist(n, vector<int>(m,0));

        bfs(mat,visited,dist);

        return dist;
    }
};
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