<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>> vec(m, vector<int>(n, -1));
        vector<vector<int>> visited(m, vector<int>(n, 0));
        queue<pair<pair<int, 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}, 0});
                    visited[i][j] = 1;
                }
            }
        }

        while (!q.empty()) {
            // pair<pair<int, int>, int> node = q.front();
            auto node = q.front();
            int i = node.first.first, j = node.first.second, dist = node.second;
            vec[i][j] = dist;
            vector<pair<int, int>> directions = {
                {0, 1}, {0, -1}, {1, 0}, {-1, 0}};
            for (auto direction : directions) {
                int x = i + direction.first, y = j + direction.second;
                if (x >= 0 && x < m && y >= 0 && y < n && visited[x][y] == 0) {
                    q.push({{x, y}, dist + 1});
                    visited[x][y] = 1;
                }
            }
            q.pop();
        }

        return vec;
    }
};
52236