Leetcode #2538: Difference Between Maximum and Minimum Price Sum
In this guide, we solve Leetcode #2538 Difference Between Maximum and Minimum Price Sum 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 initially unrooted tree with n nodes indexed from 0 to n - 1. You are given the 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, Depth-First Search, 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: n = 6, edges = [[0,1],[1,2],[1,3],[3,4],[3,5]], price = [9,8,7,6,10,5]
Output: 24
Explanation: The diagram above denotes the tree after rooting it at node 2. The first part (colored in red) shows the path with the maximum price sum. The second part (colored in blue) shows the path with the minimum price sum.
- The first path contains nodes [2,1,3,4]: the prices are [7,8,6,10], and the sum of the prices is 31.
- The second path contains the node [2] with the price [7].
The difference between the maximum and minimum price sum is 24. It can be proved that 24 is the maximum cost.
Python Solution
class Solution:
def maxOutput(self, n: int, edges: List[List[int]], price: List[int]) -> int:
def dfs(i, fa):
a, b = price[i], 0
for j in g[i]:
if j != fa:
c, d = dfs(j, i)
nonlocal ans
ans = max(ans, a + d, b + c)
a = max(a, price[i] + c)
b = max(b, price[i] + d)
return a, b
g = defaultdict(list)
for a, b in edges:
g[a].append(b)
g[b].append(a)
ans = 0
dfs(0, -1)
return ans
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
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.