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

        vector<vector<int>> ans(n, vector<int>(m, -1));

        queue<pair<int,pair<int,int>>> q;
        int visit[n][m];
        memset(visit, 0, sizeof(visit));

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

        while (!q.empty()) {
            pair<int, pair<int,int>> pr = q.front(); q.pop();

            int i = pr.second.first, j = pr.second.second;
            int d = pr.first;

            if (visit[i][j]) continue;
            visit[i][j] = 1;
            ans[i][j] = d;

            if (i-1 >= 0 && !visit[i-1][j] && mat[i-1][j]) {
                q.push({d+1, {i-1, j}});
            }
            if (j-1 >= 0 && !visit[i][j-1] && mat[i][j-1]) {
                q.push({d+1, {i, j-1}});
            }
            if (i+1 < n && !visit[i+1][j] && mat[i+1][j]) {
                q.push({d+1, {i+1, j}});
            }
            if (j+1 < m && !visit[i][j+1] && mat[i][j+1]) {
                q.push({d+1, {i, j+1}});
            }
        }

        return ans;
    }
};
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