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1443. Minimum Time to Collect All Apples in a Tree

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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
• fromi < toi
• hasApple.length == n

Solutions

DFS.

Python3

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)

Java

class Solution {
    public int minTime(int n, int[][] edges, List<Boolean> hasApple) {
        boolean[] vis = new boolean[n];
        List<Integer>[] g = new List[n];
        for (int i = 0; i < n; ++i) {
            g[i] = 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;
    }
}

C++

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;
    }
};

Go

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)
}

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