<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:

    int dx[4] = {-1, 0, +1, 0};
    int dy[4] = {0, +1, 0, -1};
    void bfs(queue<pair<int,pair<int,int>>>& q, int m, int n, vector<vector<int>>& mat, set<pair<int,int> >& st, vector<vector<int>>& res) {
        while(!q.empty()) {
            auto curr = q.front(); q.pop();
            int ci = curr.second.first;
            int cj = curr.second.second;
            int d = curr.first;
            int ni, nj, nd;
            for(int i = 0; i<4; i++) {
                ni = ci + dx[i];
                nj = cj + dy[i];
                nd = d+1;
                if(ni>=m || ni < 0 || nj >= n || nj < 0) continue;
                if(st.find({ni,nj}) == st.end()){
                    res[ni][nj] = d + 1;
                    q.push({nd,{ni,nj}});
                    st.insert({ni, nj});
                }
            }

        }
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        vector<vector<int>> res = mat;
        int m = mat.size();
        int n = mat[0].size();
        queue<pair<int,pair<int,int>>> q;
        set<pair<int,int> > st;
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(mat[i][j] == 0){
                    q.push({0,{i,j}});
                    st.insert({i,j});
                }
            }
        }

        bfs(q, m, n, mat, st, res);
        return res;
    }
};
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