<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>> bfs ( vector<vector<int>>& grid, vector <pair<int, int>> sources, vector <vector<bool>> &visited){

        int nrows = grid.size(), ncolumns = grid[0].size();
        queue <pair<int,int>> parent, child, empty;

        for ( auto source : sources ){
            int sr = source.first, sc = source.second;
            parent.push(source);
            visited[sr][sc] = true; 
        }
        
        int level = 0;
        vector<vector<int>> ans ( nrows, vector<int>(ncolumns,0));
        while ( !parent.empty() ){
            auto cv = parent.front();
            parent.pop();

            int cr = cv.first;
            int cc = cv.second;

            ans[cr][cc] = level;

            // top - left - down - right
            int dr[] = {-1, 0, 1, 0};
            int dc[] = {0, -1, 0, 1};
            for ( int i = 0 ; i < 4 ; i++ ){
                int nr = cr + dr[i];
                int nc = cc + dc[i];
                if ( nr >= 0 && nr < nrows && nc >= 0 && nc < ncolumns && !visited[nr][nc] ){
                    visited[nr][nc] = true;
                    child.push({nr,nc});
                }
            }

            if ( parent.empty() ){
                parent = child;
                child = empty;
                level++;
            }
        }

        return ans;
    }

    vector<vector<int>> updateMatrix(vector<vector<int>>& grid) {
        
        int nrows = grid.size(), ncolumns = grid[0].size();
        vector <pair<int, int>> sources; 

        for ( int ir = 0; ir < nrows ; ir++ ){
            for ( int jc = 0 ; jc < ncolumns ; jc++ ){
                if ( grid[ir][jc] == 0 ){
                    sources.push_back({ir,jc});
                }
            }
        }

        vector <vector<bool>> visited (nrows, vector<bool> (ncolumns, false) );
        vector <vector<int>> ans = bfs ( grid, sources, visited );

        return ans;

    }
};
44898