#2538

Hard

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Difference Between Maximum and Minimum Price Sum

Compute the maximum difference between any path price sum in a tree using binary-tree traversal and state tracking efficiently.

array dynamic programming tree depth first search

Problem Statement

You are given an undirected tree with n nodes labeled from 0 to n - 1, described by a list of edges. Each node has an associated price given in an array, and a path's price sum is the sum of prices along its nodes. Calculate the maximum difference between any path's price sum and the smallest path price sum in the tree.

Design an algorithm that efficiently traverses the tree to track all possible path sums, leveraging binary-tree traversal and state tracking. Consider the minimum path sum as the price of a single node and ensure the solution scales for up to 10^5 nodes.

Examples

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

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

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 <= ai, bi <= n - 1
edges represents a valid tree.
price.length == n
1 <= price[i] <= 105

Solution Approach

Root the Tree and Initialize State

Choose an arbitrary node as the root and perform a depth-first traversal to initialize cumulative price sums for each subtree. Maintain arrays or dictionaries to track both maximum and minimum path sums from each node, as this problem requires careful state tracking along tree edges.

Depth-First Search to Compute Path Sums

Recursively traverse each child node while updating the current path sum. At each step, record the maximum path sum reachable from the node and the minimum path sum considering single-node paths. This ensures the difference can be computed correctly and avoids double counting paths.

Calculate Maximum Difference

After DFS, iterate through all nodes' tracked states to find the maximum path sum and the minimum single-node price sum. The final answer is the difference between these two values, directly solving the problem with precise state tracking and leveraging dynamic programming on trees.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) due to a single DFS traversal through all nodes. Space complexity is O(n) for storing subtree states and recursion stack, which scales with the tree size.

What Interviewers Usually Probe

Expect efficient handling of trees with up to 10^5 nodes, highlighting binary-tree traversal mastery.
Demonstrate understanding of state tracking and dynamic programming across tree edges.
Recognize that the minimum price sum is always a single node, which guides optimal traversal.

Common Pitfalls or Variants

Common pitfalls

Forgetting to consider single-node paths as minimum sum, leading to incorrect difference.
Recomputing path sums for overlapping subtrees instead of tracking state, causing inefficiency.
Incorrectly summing path prices without maintaining subtree maximums, missing the global maximum difference.

Follow-up variants

Compute difference between maximum and minimum weighted path sums with additional edge costs.
Extend the problem to handle multiple disconnected trees (forest) and report maximum differences per tree.
Modify node prices dynamically and recalculate the maximum difference efficiently for online queries.

FAQ

What is the main pattern used in Difference Between Maximum and Minimum Price Sum?

The problem relies on binary-tree traversal and state tracking to calculate path sums efficiently.

How do I find the minimum path sum?

The minimum path sum is always the price of a single node, typically chosen as the root or any leaf.

Can this solution handle large n values?

Yes, using DFS with state tracking ensures O(n) time and space complexity, suitable for up to 10^5 nodes.

Do I need dynamic programming for this problem?

Yes, dynamic programming on tree edges allows tracking maximum path sums without redundant computations.

How is the final difference calculated?

Subtract the minimum single-node price from the maximum path sum recorded during DFS traversal to get the answer.

terminal

Solution

Solution 1

#### Python3

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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

Continue Practicing

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