Leetcode #2603: Collect Coins in a Tree
In this guide, we solve Leetcode #2603 Collect Coins 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
There exists an undirected and unrooted tree with n nodes indexed from 0 to n - 1. You are given an integer n and a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Graph, Topological Sort, Array
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: coins = [1,0,0,0,0,1], edges = [[0,1],[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation: Start at vertex 2, collect the coin at vertex 0, move to vertex 3, collect the coin at vertex 5 then move back to vertex 2.
Python Solution
class Solution:
def collectTheCoins(self, coins: List[int], edges: List[List[int]]) -> int:
g = defaultdict(set)
for a, b in edges:
g[a].add(b)
g[b].add(a)
n = len(coins)
q = deque(i for i in range(n) if len(g[i]) == 1 and coins[i] == 0)
while q:
i = q.popleft()
for j in g[i]:
g[j].remove(i)
if coins[j] == 0 and len(g[j]) == 1:
q.append(j)
g[i].clear()
for k in range(2):
q = [i for i in range(n) if len(g[i]) == 1]
for i in q:
for j in g[i]:
g[j].remove(i)
g[i].clear()
return sum(len(g[a]) > 0 and len(g[b]) > 0 for a, b in edges) * 2
Complexity
The time complexity is and the space complexity is , where is the number of nodes. The space complexity is , where is the number of nodes.
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.