Leetcode #1443: Minimum Time to Collect All Apples in a Tree
In this guide, we solve Leetcode #1443 Minimum Time to Collect All Apples in a Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.