<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 height = mat.size();
        int length = mat[0].size();
        queue<pair<int,pair<int,int>>> q;
        unordered_set<int> visited;
        for (int i = 0; i < mat.size(); i++) {
            for (int j = 0; j < mat[0].size(); j++) {
                if (mat[i][j] == 0) {
                    q.push({0,{i,j}});
                    visited.insert(i*length+j);
                }
            }
        }
        while (!q.empty()) {
            pair<int,pair<int,int>> curr = q.front();
            q.pop();
            int dist = curr.first;
            int r = curr.second.first;
            int c = curr.second.second;
            if (r-1 >= 0 && visited.count((r-1)*length+c) <= 0) {
                visited.insert((r-1)*length+c);
                mat[r-1][c] = dist+1;
                q.push({dist+1,{r-1,c}});
            } 
            if (c+1 < length && visited.count(r*length+(c+1)) <= 0) {
                visited.insert(r*length+(c+1));
                mat[r][c+1] = dist+1;
                q.push({dist+1,{r,c+1}});
            } 
            if (r+1 < height && visited.count((r+1)*length+c) <= 0) {
                visited.insert((r+1)*length+c);
                mat[r+1][c] = dist+1;
                q.push({dist+1,{r+1,c}});
            } 
            if (c-1 >= 0 && visited.count(r*length+(c-1)) <= 0) {
                visited.insert(r*length+(c-1));
                mat[r][c-1] = dist+1;
                q.push({dist+1,{r,c-1}});
            } 
        }
        return mat;
    }
};
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