Welcome to Subscribe On Youtube

1443. Minimum Time to Collect All Apples in a Tree

Description

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
• hasApple.length == n

Solutions

DFS.

• Java
• C++
• Python
• Go

• class Solution {
      public int minTime(int n, int[][] edges, List<Boolean> hasApple) {
          boolean[] vis = new boolean[n];
          List<Integer>[] g = new List[n];
          Arrays.setAll(g, k -> new ArrayList<>());
          for (int[] e : edges) {
              int u = e[0], v = e[1];
              g[u].add(v);
              g[v].add(u);
          }
          return dfs(0, 0, g, hasApple, vis);
      }
  
      private int dfs(int u, int cost, List<Integer>[] g, List<Boolean> hasApple, boolean[] vis) {
          if (vis[u]) {
              return 0;
          }
          vis[u] = true;
          int nxtCost = 0;
          for (int v : g[u]) {
              nxtCost += dfs(v, 2, g, hasApple, vis);
          }
          if (!hasApple.get(u) && nxtCost == 0) {
              return 0;
          }
          return cost + nxtCost;
      }
  }
  

• class Solution {
  public:
      int minTime(int n, vector<vector<int>>& edges, vector<bool>& hasApple) {
          vector<bool> vis(n);
          vector<vector<int>> g(n);
          for (auto& e : edges) {
              int u = e[0], v = e[1];
              g[u].push_back(v);
              g[v].push_back(u);
          }
          return dfs(0, 0, g, hasApple, vis);
      }
  
      int dfs(int u, int cost, vector<vector<int>>& g, vector<bool>& hasApple, vector<bool>& vis) {
          if (vis[u]) return 0;
          vis[u] = true;
          int nxt = 0;
          for (int& v : g[u]) nxt += dfs(v, 2, g, hasApple, vis);
          if (!hasApple[u] && !nxt) return 0;
          return cost + nxt;
      }
  };
  

• class Solution:
      def minTime(self, n: int, edges: List[List[int]], hasApple: List[bool]) -> int:
          def dfs(u, cost):
              if vis[u]:
                  return 0
              vis[u] = True
              nxt_cost = 0
              for v in g[u]:
                  nxt_cost += dfs(v, 2)
              if not hasApple[u] and nxt_cost == 0:
                  return 0
              return cost + nxt_cost
  
          g = defaultdict(list)
          for u, v in edges:
              g[u].append(v)
              g[v].append(u)
          vis = [False] * n
          return dfs(0, 0)
  
  

• func minTime(n int, edges [][]int, hasApple []bool) int {
  	vis := make([]bool, n)
  	g := make([][]int, n)
  	for _, e := range edges {
  		u, v := e[0], e[1]
  		g[u] = append(g[u], v)
  		g[v] = append(g[v], u)
  	}
  	var dfs func(int, int) int
  	dfs = func(u, cost int) int {
  		if vis[u] {
  			return 0
  		}
  		vis[u] = true
  		nxt := 0
  		for _, v := range g[u] {
  			nxt += dfs(v, 2)
  		}
  		if !hasApple[u] && nxt == 0 {
  			return 0
  		}
  		return cost + nxt
  	}
  	return dfs(0, 0)
  }
  

All Problems

All Solutions