How do you solve Minimum Edge Weight Equilibrium Queries in a Tree on LeetCode?

Minimum Edge Weight Equilibrium Queries in a Tree (LeetCode #2846) 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: Understanding tree structure is crucial for efficient traversal and pathfinding.

What is the time complexity of Minimum Edge Weight Equilibrium Queries in a Tree?

The optimal solution for Minimum Edge Weight Equilibrium Queries in a Tree runs in O(n) time and uses O(n) space. This complexity is due to the single DFS traversal to build the frequency map, making it efficient for multiple queries.

What is the brute force solution for Minimum Edge Weight Equilibrium Queries in a Tree?

The brute force approach for Minimum Edge Weight Equilibrium Queries in a Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. For each query, we can find the path between the two nodes and count how many edges need to be changed to make all their weights equal. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Minimum Edge Weight Equilibrium Queries in a Tree?

Always clarify the problem constraints and edge cases before diving into the solution. Discuss your thought process and approach with the interviewer to demonstrate your understanding. Practice explaining your solution clearly and concisely, as communication is key in interviews.

What are common mistakes when solving Minimum Edge Weight Equilibrium Queries in a Tree?

Not considering the independent nature of queries, leading to incorrect assumptions about edge weights. Failing to optimize the pathfinding process, resulting in unnecessary complexity.

#2846

Minimum Edge Weight Equilibrium Queries in a Tree

Hard

ArrayDynamic ProgrammingBit ManipulationTreeDepth-First SearchDFS for tree traversalFrequency counting with Hash Maps

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)

By rooting the tree and using a frequency map to count edge weights along paths, we can efficiently determine the minimum changes needed for each query.

⚙️

Algorithm

3 steps

1Step 1: Root the tree at an arbitrary node (e.g., node 0).
2Step 2: Create a frequency map (2D array) to store the count of each edge weight for paths from the root to each node.
3Step 3: For each query, use the precomputed frequency map to determine the maximum frequency of edge weights along the path and calculate the required changes.

solution.py22 lines

1def min_operations_optimal(n, edges, queries):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for u, v, w in edges:
5        graph[u].append((v, w))
6        graph[v].append((u, w))
7
8    freq = [defaultdict(int) for _ in range(n)]
9    def dfs(node, parent):
10        for neighbor, weight in graph[node]:
11            if neighbor != parent:
12                freq[neighbor][weight] += 1
13                for w, count in freq[node].items():
14                    freq[neighbor][w] += count
15                dfs(neighbor, node)
16
17    dfs(0, -1)
18    results = []
19    for a, b in queries:
20        max_freq = max(freq[a].values(), default=0) + max(freq[b].values(), default=0)
21        results.append((len(freq[a]) + len(freq[b])) - max_freq)
22    return results

ℹ

Complexity note: This complexity is due to the single DFS traversal to build the frequency map, making it efficient for multiple queries.

1Understanding tree structure is crucial for efficient traversal and pathfinding.
2Using frequency maps allows for quick calculations of edge weights along paths.

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