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.

4018. Total Sum of Interaction Cost in Tree Groups II - Practice Online

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] = [u_i, v_i] indicates an undirected edge between nodes u_i and v_i.

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:

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