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[1/11] 1443. Minimum Time to Collect All Apples in a Tree

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Given an undirected tree consisting of n vertices numbered from 0 to n-1, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. Return the minimum time in seconds you have to spend to collect all apples in the tree, starting at vertex 0 and coming back to this vertex.

The edges of the undirected tree are given in the array edges, where edges[i] = [ai, bi] means that exists an edge connecting the vertices ai and bi. Additionally, there is a boolean array hasApple, where hasApple[i] = true means that vertex i has an apple; otherwise, it does not have any apple.

 

Example 1:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,true,true,false]
Output: 8 
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.  

Example 2:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,false,true,false]
Output: 6
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.  

Example 3:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,false,false,false,false,false]
Output: 0

 

Constraints:

  • 1 <= n <= 105
  • edges.length == n - 1
  • edges[i].length == 2
  • 0 <= ai < bi <= n - 1
  • fromi < toi
  • hasApple.length == n
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from collections import defaultdict
class Solution:
    def minTime(self, n: int, edges: List[List[int]], hasApple: List[bool]) -> int:
        graph = defaultdict(list)
        for n1, n2 in edges:
            graph[n1].append(n2)
            graph[n2].append(n1)
        
        need_visit = set()

        def dfs(prev, curr):
            res = False
            for nxt in graph[curr]:
                
                if nxt == prev: continue
                temp = dfs(curr, nxt)
                res = res or temp
            
            res = res or hasApple[curr]
            if res: need_visit.add(curr)
            return res
        
        dfs(-1,0)
        return max(2*len(need_visit) - 2,0)

        
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