Leetcode #2920: Maximum Points After Collecting Coins From All Nodes
In this guide, we solve Leetcode #2920 Maximum Points After Collecting Coins From All Nodes 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 tree rooted at node 0 with n nodes labeled from 0 to n - 1. You are given 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: Bit Manipulation, Tree, Depth-First Search, Memoization, Array, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: edges = [[0,1],[1,2],[2,3]], coins = [10,10,3,3], k = 5
Output: 11
Explanation:
Collect all the coins from node 0 using the first way. Total points = 10 - 5 = 5.
Collect all the coins from node 1 using the first way. Total points = 5 + (10 - 5) = 10.
Collect all the coins from node 2 using the second way so coins left at node 3 will be floor(3 / 2) = 1. Total points = 10 + floor(3 / 2) = 11.
Collect all the coins from node 3 using the second way. Total points = 11 + floor(1 / 2) = 11.
It can be shown that the maximum points we can get after collecting coins from all the nodes is 11.
Python Solution
class Solution:
def maximumPoints(self, edges: List[List[int]], coins: List[int], k: int) -> int:
def dfs(i: int, fa: int, j: int) -> int:
a = (coins[i] >> j) - k
b = coins[i] >> (j + 1)
for c in g[i]:
if c != fa:
a += dfs(c, i, j)
if j < 14:
b += dfs(c, i, j + 1)
return max(a, b)
n = len(coins)
g = [[] for _ in range(n)]
for a, b in edges:
g[a].append(b)
g[b].append(a)
ans = dfs(0, -1, 0)
dfs.cache_clear()
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.