Leetcode #2925: Maximum Score After Applying Operations on a Tree
In this guide, we solve Leetcode #2925 Maximum Score After Applying Operations on 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 is an undirected tree with n nodes labeled from 0 to n - 1, and rooted at node 0. 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: Medium
- Premium: No
- Tags: Tree, Depth-First Search, 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],[0,2],[0,3],[2,4],[4,5]], values = [5,2,5,2,1,1]
Output: 11
Explanation: We can choose nodes 1, 2, 3, 4, and 5. The value of the root is non-zero. Hence, the sum of values on the path from the root to any leaf is different than zero. Therefore, the tree is healthy and the score is values[1] + values[2] + values[3] + values[4] + values[5] = 11.
It can be shown that 11 is the maximum score obtainable after any number of operations on the tree.
Python Solution
class Solution:
def maximumScoreAfterOperations(
self, edges: List[List[int]], values: List[int]
) -> int:
def dfs(i: int, fa: int = -1) -> (int, int):
a = b = 0
leaf = True
for j in g[i]:
if j != fa:
leaf = False
aa, bb = dfs(j, i)
a += aa
b += bb
if leaf:
return values[i], 0
return values[i] + a, max(values[i] + b, a)
g = [[] for _ in range(len(values))]
for a, b in edges:
g[a].append(b)
g[b].append(a)
return dfs(0)[1]
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.