How do you solve Total Sum of Interaction Cost in Tree Groups on LeetCode?

Total Sum of Interaction Cost in Tree Groups (LeetCode #3786) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Count nodes in each group efficiently using DFS.

What is the time complexity of Total Sum of Interaction Cost in Tree Groups?

The optimal solution for Total Sum of Interaction Cost in Tree Groups runs in O(n) time and uses O(n) space. The DFS traverses each node once, leading to linear time complexity, while storing counts requires O(n) space.

What is the brute force solution for Total Sum of Interaction Cost in Tree Groups?

The brute force approach for Total Sum of Interaction Cost in Tree Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. Calculate the interaction cost for every pair of nodes in the same group by traversing the tree. This involves finding the path length for each pair, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Total Sum of Interaction Cost in Tree Groups?

Total Sum of Interaction Cost in Tree Groups uses the following concepts: Array, Tree, Depth-First Search. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Total Sum of Interaction Cost in Tree Groups?

Explain your thought process clearly. Discuss edge cases and constraints. Practice DFS and tree traversal problems.

What are common mistakes when solving Total Sum of Interaction Cost in Tree Groups?

Not considering all pairs correctly. Forgetting to account for unordered pairs.

#3786

Total Sum of Interaction Cost in Tree Groups

Hard

ArrayTreeDepth-First SearchHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use DFS to count nodes in each subtree for each group. For each edge, calculate the contribution to the interaction cost based on the number of nodes in the subtrees.

⚙️

Algorithm

3 steps

1Step 1: Build the tree from edges and perform a DFS to count nodes in each group for each subtree.
2Step 2: For each edge, calculate the contribution as subtree_count * (total_count - subtree_count).
3Step 3: Sum all contributions to get the total interaction cost.

solution.py22 lines

1def total_interaction_cost(n, edges, group):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for u, v in edges:
5        graph[u].append(v)
6        graph[v].append(u)
7    group_count = defaultdict(int)
8    visited = [False] * n
9    total_cost = 0
10    def dfs(node):
11        nonlocal total_cost
12        visited[node] = True
13        count = 1
14        for neighbor in graph[node]:
15            if not visited[neighbor]:
16                subtree_count = dfs(neighbor)
17                total_cost += subtree_count * (group_count[group[node]] - subtree_count)
18                count += subtree_count
19        group_count[group[node]] += count
20        return count
21    dfs(0)
22    return total_cost

ℹ

Complexity note: The DFS traverses each node once, leading to linear time complexity, while storing counts requires O(n) space.

1Count nodes in each group efficiently using DFS.
2Understand how paths in trees relate to edge contributions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.