<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();

        int max_dist = max(m, n);

        vector<vector<int>> mat_dist(m, vector<int>(n, max_dist));

        queue<pair<int,int>> q;

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

        while (!q.empty()) {
            int p = q.size();
            cout << "p = " << p << endl;

            for (int k = 0; k < p; k++) {
                pair<int, int> cell = q.front();
                int x = cell.first, y = cell.second;
                q.pop();

                // skip if already updated from default (i.e. visited)
                if (mat_dist[x][y] != max_dist)
                    continue;

                int min_dist =  (mat[x][y] == 0) ? 0 : max_dist;
                if (x > 0)   {
                    min_dist = min(min_dist, mat_dist[x-1][y] + 1);
                    q.push({x-1, y});
                }
                if (x < m-1) {
                    min_dist = min(min_dist, mat_dist[x+1][y] + 1);
                    q.push({x+1, y});
                }
                if (y > 0) {
                    min_dist = min(min_dist, mat_dist[x][y-1] + 1);
                    q.push({x, y-1});
                }
                if (y < n-1) {
                    min_dist = min(min_dist, mat_dist[x][y+1] + 1);
                    q.push({x, y+1});
                }

                mat_dist[x][y] = min_dist;

                
            }
        }

        return mat_dist;


    }
};
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