Is Subtree Inversion Sum II easy or hard?

Subtree Inversion Sum II is considered a hard problem because it combines multiple advanced ideas: tree traversal, inversion counting, and efficient data structure merging. Recognizing that naive subtree recomputation is too slow and switching to DSU-on-tree or Fenwick-based aggregation is the key difficulty.

Subtree Inversion Sum II Python/Java solution

Python implementations usually use coordinate compression with a Fenwick Tree or the small-to-large merging technique using sorted containers. Java solutions often rely on TreeMap, Fenwick Tree arrays, or custom DSU-on-tree implementations. All optimal versions run in O(n log n) time.

How to solve Subtree Inversion Sum II in O(n log n)?

Perform a DFS over the tree and maintain a structure storing values from the processed subtree. Using small-to-large merging ensures each element is moved only a limited number of times. When inserting a value, query how many existing values are greater using a balanced BST or Fenwick Tree, which counts new inversion pairs in logarithmic time.

What is the best approach for Subtree Inversion Sum II?

The most practical solution uses DFS with small-to-large merging (DSU on tree). Each subtree maintains an ordered structure of values, and smaller child structures are merged into the largest one while counting how many existing values create inversions. This reduces repeated work and achieves O(n log n) time with O(n) space.

Is Subtree Inversion Sum II asked at Google/Amazon/Meta?

Problems combining subtree processing, inversion counting, and ordered data structures are common in interviews at companies like Google, Amazon, and Meta. Variants appear as tree queries, DSU-on-tree tasks, or Fenwick Tree applications where candidates must aggregate subtree statistics efficiently.

What data structure is used in Subtree Inversion Sum II?

Typical solutions rely on ordered structures such as a multiset, balanced binary search tree, or Fenwick Tree after coordinate compression. These structures allow fast insertion and queries for how many values are greater or smaller, which is essential for counting inversion pairs.

What is the time complexity of Subtree Inversion Sum II?

The optimal solutions run in O(n log n) time. Techniques such as DSU on tree or Euler tour combined with a Fenwick Tree allow each node value to be inserted and queried logarithmically. A naive approach that recomputes inversion counts for every subtree can degrade to O(n^2 log n).

3949. Subtree Inversion Sum II - Practice Online

You are given an undirected tree rooted at node 0, with n nodes numbered from 0 to n - 1. The tree is represented by a 2D integer array edges of length n - 1, where edges[i] = [u_i, v_i] indicates an edge between nodes u_i and v_i.

You are also given an integer array nums of length n, where nums[i] represents the value at node i, and an integer k.

You may perform inversion operations on a subset of nodes subject to the following rules:

Subtree Inversion Operation:
- When you invert a node, every value in the subtree rooted at that node is multiplied by -1.
Distance Constraint on Inversions:
- You may only invert a node if it is “sufficiently far” from any other inverted node.
- If you invert two nodes a and b, the distance (the number of edges on the unique path between them) must be at least k.

Return the maximum possible sum of the tree’s node values after applying inversion operations.

Example 1:

Input: edges = [[0,1],[0,2],[0,3],[1,4],[1,5]], nums = [1,0,-10,3,4,5], k = 2

Output: 23

Explanation:

After inverting the subtree rooted at node 2, the maximum sum becomes 1 + 0 + 10 + 3 + 4 + 5 = 23.

Example 2:

Input: edges = [[0,1],[1,2]], nums = [5,-10,-10], k = 1

Output: 25

Explanation:

After inverting the subtree rooted at node 1, the maximum sum becomes 5 + 10 + 10 = 25.

Example 3:

Input: edges = [[0,1],[0,2]], nums = [1,-5,-6], k = 2

Output: 12

Explanation:

After inverting the subtrees rooted at nodes 1 and 2, nums = [1, 5, 6].
This is valid because nodes 1 and 2 are two edges apart (1 → 0 and 0 → 2), which is at least k.
The maximum sum is 1 + 5 + 6 = 12.

Example 4:

Input: edges = [[0,1],[0,2]], nums = [1,-5,-6], k = 3

Output: 10

Explanation:

After inverting the subtree rooted at nodes 0, nums = [-1, 5, 6].
The maximum sum is (-1) + 5 + 6 = 10.
Note that we cannot invert nodes 1 and 2 because their distance is 2 < k = 3.

Constraints:

nums.length == n
edges.length == n - 1
2 <= n <= 5 * 10⁴
edges[i].length == 2
0 <= edges[i][0], edges[i][1] < n
-4 * 10⁴ <= nums[i] <= 4 * 10⁴
1 <= k <= 50
It is guaranteed that edges forms a tree.