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:
![](https://spcdn.pages.dev/leetcode/problems/1443.Minimum%20Time%20to%20Collect%20All%20Apples%20in%20a%20Tree/images/min_time_collect_apple_1.png)
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:
![](https://spcdn.pages.dev/leetcode/problems/1443.Minimum%20Time%20to%20Collect%20All%20Apples%20in%20a%20Tree/images/min_time_collect_apple_2.png)
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#
Solution 1#
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| 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)
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| 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;
}
}
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| 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;
}
};
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| 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)
}
|