LeetCode #2538 — HARD

Difference Between Maximum and Minimum Price Sum

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree.

Each node has an associated price. You are given an integer array price, where price[i] is the price of the i^th node.

The price sum of a given path is the sum of the prices of all nodes lying on that path.

The tree can be rooted at any node root of your choice. The incurred cost after choosing root is the difference between the maximum and minimum price sum amongst all paths starting at root.

Return the maximum possible cost amongst all possible root choices.

Example 1:

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.

Example 2:

Input: n = 3, edges = [[0,1],[1,2]], price = [1,1,1]
Output: 2
Explanation: The diagram above denotes the tree after rooting it at node 0. 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 [0,1,2]: the prices are [1,1,1], and the sum of the prices is 3.
- The second path contains node [0] with a price [1].
The difference between the maximum and minimum price sum is 2. It can be proved that 2 is the maximum cost.

Constraints:

1 <= n <= 10⁵
edges.length == n - 1
0 <= a_i, b_i <= n - 1
edges represents a valid tree.
price.length == n
1 <= price[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: 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. Each node has an associated price. You are given an integer array price, where price[i] is the price of the ith node. The price sum of a given path is the sum of the prices of all nodes lying on that path. The tree can be rooted at any node root of your choice. The incurred cost after choosing root is the difference between the maximum and minimum price sum amongst all paths starting at root. Return the maximum possible cost amongst all possible root choices.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Tree

Example 1

6
[[0,1],[1,2],[1,3],[3,4],[3,5]]
[9,8,7,6,10,5]

Example 2

3
[[0,1],[1,2]]
[1,1,1]

Core Insight

What unlocks the optimal approach

The minimum price sum is always the price of a rooted node.
Let’s root the tree at vertex 0 and find the answer from this perspective.
In the optimal answer maximum price is the sum of the prices of nodes on the path from “u” to “v” where either “u” or “v” is the parent of the second one or neither is a parent of the second one.
The first case is easy to find. For the second case, notice that in the optimal path, “u” and “v” are both leaves. Then we can use dynamic programming to find such a path.
Let DP(v,1) denote “the maximum price sum from node v to leaf, where v is a parent of that leaf” and let DP(v,0) denote “the maximum price sum from node v to leaf, where v is a parent of that leaf - price[leaf]”. Then the answer is maximum of DP(u,0) + DP(v,1) + price[parent] where u, v are directly connected to vertex “parent”.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2538: Difference Between Maximum and Minimum Price Sum
class Solution {
    private List<Integer>[] g;
    private long ans;
    private int[] price;

    public long maxOutput(int n, int[][] edges, int[] price) {
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        this.price = price;
        dfs(0, -1);
        return ans;
    }

    private long[] dfs(int i, int fa) {
        long a = price[i], b = 0;
        for (int j : g[i]) {
            if (j != fa) {
                var e = dfs(j, i);
                long c = e[0], d = e[1];
                ans = Math.max(ans, Math.max(a + d, b + c));
                a = Math.max(a, price[i] + c);
                b = Math.max(b, price[i] + d);
            }
        }
        return new long[] {a, b};
    }
}

// Accepted solution for LeetCode #2538: Difference Between Maximum and Minimum Price Sum
func maxOutput(n int, edges [][]int, price []int) int64 {
	g := make([][]int, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	type pair struct{ a, b int }
	ans := 0
	var dfs func(i, fa int) pair
	dfs = func(i, fa int) pair {
		a, b := price[i], 0
		for _, j := range g[i] {
			if j != fa {
				e := dfs(j, i)
				c, d := e.a, e.b
				ans = max(ans, max(a+d, b+c))
				a = max(a, price[i]+c)
				b = max(b, price[i]+d)
			}
		}
		return pair{a, b}
	}
	dfs(0, -1)
	return int64(ans)
}

# Accepted solution for LeetCode #2538: Difference Between Maximum and Minimum Price Sum
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

// Accepted solution for LeetCode #2538: Difference Between Maximum and Minimum Price Sum
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2538: Difference Between Maximum and Minimum Price Sum
// class Solution {
//     private List<Integer>[] g;
//     private long ans;
//     private int[] price;
// 
//     public long maxOutput(int n, int[][] edges, int[] price) {
//         g = new List[n];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (var e : edges) {
//             int a = e[0], b = e[1];
//             g[a].add(b);
//             g[b].add(a);
//         }
//         this.price = price;
//         dfs(0, -1);
//         return ans;
//     }
// 
//     private long[] dfs(int i, int fa) {
//         long a = price[i], b = 0;
//         for (int j : g[i]) {
//             if (j != fa) {
//                 var e = dfs(j, i);
//                 long c = e[0], d = e[1];
//                 ans = Math.max(ans, Math.max(a + d, b + c));
//                 a = Math.max(a, price[i] + c);
//                 b = Math.max(b, price[i] + d);
//             }
//         }
//         return new long[] {a, b};
//     }
// }

// Accepted solution for LeetCode #2538: Difference Between Maximum and Minimum Price Sum
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2538: Difference Between Maximum and Minimum Price Sum
// class Solution {
//     private List<Integer>[] g;
//     private long ans;
//     private int[] price;
// 
//     public long maxOutput(int n, int[][] edges, int[] price) {
//         g = new List[n];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (var e : edges) {
//             int a = e[0], b = e[1];
//             g[a].add(b);
//             g[b].add(a);
//         }
//         this.price = price;
//         dfs(0, -1);
//         return ans;
//     }
// 
//     private long[] dfs(int i, int fa) {
//         long a = price[i], b = 0;
//         for (int j : g[i]) {
//             if (j != fa) {
//                 var e = dfs(j, i);
//                 long c = e[0], d = e[1];
//                 ans = Math.max(ans, Math.max(a + d, b + c));
//                 a = Math.max(a, price[i] + c);
//                 b = Math.max(b, price[i] + d);
//             }
//         }
//         return new long[] {a, b};
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.