<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) {
        unordered_set<int> v;
        queue<pair<int,int>>q;
        int r = mat.size();
        int c = mat[0].size();
        for(int i=0;i<r;i++){
            for(int j =0;j<c;j++){
                if(mat[i][j] == 0){
                    q.push({i,j});
                    v.insert(i*c + j);
                }
            }
        }
        vector<pair<int,int>> next = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        while(!q.empty()){
            pair<int,int> top = q.front();
            q.pop();
            int row = top.first;
            int col = top.second;
            for(auto &na:next){
                int nr = row + na.first;
                int nc = col + na.second;
                if(nr >= 0 && nc >= 0 && nr < r && nc < c && !v.count(nr * c + nc)){
                    mat[nr][nc] = mat[row][col]+1;
                    v.insert(nr *c + nc);
                    q.push({nr,nc});
                }
            }
        }
        return mat;
    }
};
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