<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 m, n;
    vector<vector<int>> dp;
    
    void bfs(vector<pair<int, int>>& mp){
        queue<pair<int, int>> q;
        set<pair<int, int>> st;
        vector<pair<int, int>> dir {{1,0},{-1,0},{0,1},{0,-1}};
        
        for(auto i: mp){
            q.push(i);
            st.insert(i);
        }
        
        while(!q.empty()){
            pair<int, int> temp = q.front();
            q.pop();
            for(auto i: dir){
                if(temp.first+i.first <0 || temp.first+i.first >=m) continue;
                if(temp.second+i.second <0 || temp.second+i.second >=n) continue;
                
                dp[temp.first+i.first][temp.second+i.second] = min(dp[temp.first+i.first][temp.second+i.second], dp[temp.first][temp.second]+1);
                if(st.find({temp.first+i.first, temp.second+i.second})==st.end()){
                    st.insert({temp.first+i.first, temp.second+i.second});
                    q.push({temp.first+i.first, temp.second+i.second});
                }
            }            
        }
    }
    
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        m=mat.size(); n=mat[0].size();
        dp.resize(m, vector<int>(n, 1e4));
        
        vector<pair<int, int>> mp;
        
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(!mat[i][j]){
                    mp.push_back({i, j});
                    dp[i][j]=0;
                }
            }
        }
        
        bfs(mp);
        return dp;
    }
};
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