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Total Sum of Interaction Cost in Tree Groups II - Solution & Explanation

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Problem Statement

You are given an integer n and 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] = [ui, vi] indicates an undirected edge between nodes ui and vi.

You are also given an integer array group of length n, where group[i] denotes the group label assigned to node i.

  • Two nodes u and v belong to the same group if and only if group[u] == group[v].
  • The interaction cost between two nodes is the shortest distance between them in the tree.

Return the sum of interaction costs over all pairs of node indices (u, v) such that 0 <= u < v < n and group[u] == group[v].

The shortest distance between two nodes is the number of edges on the unique path connecting them in the tree.

 

Example 1:

Input: n = 3, edges = [[0,1],[1,2]], group = [1,1,1]

Output: 4

Explanation:

All nodes belong to group 1. The interaction costs between the pairs of nodes are:

  • Nodes [0, 1]: 1
  • Nodes [1, 2]: 1
  • Nodes [0, 2]: 2

Thus, the total interaction cost is 1 + 1 + 2 = 4.

Example 2:

Input: n = 3, edges = [[0,1],[1,2]], group = [3,2,3]

Output: 2

Explanation:

  • Nodes 0 and 2 belong to group 3. The interaction cost between this pair is 2.
  • Node 1 belongs to a different group and forms no valid pair. Therefore, the total interaction cost is 2.

Example 3:

Input: n = 4, edges = [[0,1],[0,2],[0,3]], group = [1,1,4,4]

Output: 3

Explanation:

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Nodes belonging to the same groups and their interaction costs are:

  • Group 1: Nodes [0, 1]: 1
  • Group 4: Nodes [2, 3]: 2

Thus, the total interaction cost is 1 + 2 = 3.

Example 4:

Input: n = 2, edges = [[0,1]], group = [1,2]

Output: 0

Explanation:

All nodes belong to different groups and there are no valid pairs. Therefore, the total interaction cost is 0.

 

Constraints:

  • 1 <= n <= 105
  • edges.length == n - 1
  • edges[i] = [ui, vi]
  • 0 <= ui, vi <= n - 1
  • group.length == n
  • 1 <= group[i] <= n
  • The input is generated such that edges represents a valid tree.

Approach Overview

Problem Overview: Given a tree with nodes partitioned into groups, you need to find the total sum of interaction costs between all pairs of distinct groups, where the cost between two groups is defined as the sum of distances from every node in one group to every node in the other group.

Approach 1: Brute Force (O(n^2) time, O(n) space)

For each pair of groups, compute the sum of distances between all nodes in those groups by running BFS/DFS from each node in one group and summing distances to nodes in the other. This repeats work for every pair and every node, leading to quadratic complexity. It's straightforward but only feasible for small trees or few groups.

Approach 2: Edge Contribution with DFS (O(n) time, O(n) space)

The key insight is to count how many times each edge contributes to the total cost. For an edge connecting a subtree to the rest of the tree, it contributes exactly size_subtree * (total_nodes - size_subtree) times for each pair of nodes across that edge. To incorporate group costs, you need to track how many nodes from each group lie on each side of the edge. A single DFS computes subtree sizes and group counts per subtree, then you accumulate contributions for every edge. This reduces the problem to a single traversal and avoids pairwise enumeration.

Recommended for interviews: Interviewers expect the edge-contribution approach because it demonstrates deep understanding of tree properties and combinatorial counting. The brute force shows you can model the problem, but the optimal solution proves you can optimize using structural insights. Always start with brute force to clarify the logic, then pivot to the linear solution.

This problem builds on tree traversal and depth-first search, and also touches on combinatorics.

Solutions for this problem are being prepared.

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force (Pairwise BFS)O(n^2)O(n)Small trees or when groups are few
Edge Contribution with DFSO(n)O(n)General case and large inputs

Frequently Asked Questions

Is Total Sum of Interaction Cost in Tree Groups II easy or hard?
It is rated Hard because it requires deriving a non-obvious edge-contribution formula and handling multiple groups efficiently. The brute force is easy but fails on large inputs.
Total Sum of Interaction Cost in Tree Groups II Python/Java solution
Implement a recursive DFS that returns subtree size and group counts. Accumulate edge contributions using precomputed total counts per group. Both Python and Java solutions follow this pattern with iterative or recursive DFS.
How to solve Total Sum of Interaction Cost in Tree Groups II in O(n)?
Perform a DFS from any root to compute subtree sizes and per-group counts for each subtree. For each edge, its contribution is calculated as (nodes from group A on one side * nodes from group B on the other side) summed over all group pairs, which simplifies to total pairs across that edge.
What is the best approach for Total Sum of Interaction Cost in Tree Groups II?
The optimal approach uses a single DFS to compute subtree sizes and group counts, then accumulates edge contributions in O(n) time and O(n) space. This avoids quadratic pairwise computations.
Is Total Sum of Interaction Cost in Tree Groups II asked at Google/Amazon/Meta?
This problem tests advanced tree traversal and combinatorial counting, which are common topics at top tech companies like Google, Amazon, and Meta. Expect similar problems in onsite interviews.
What data structure is used in Total Sum of Interaction Cost in Tree Groups II?
The primary data structure is a tree represented as an adjacency list. You also use arrays to store subtree sizes and group counts during DFS.
What is the time complexity of Total Sum of Interaction Cost in Tree Groups II?
The optimal solution runs in O(n) time, where n is the number of nodes in the tree. The brute force approach takes O(n^2) time.

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