How do you solve Maximize Sum of Weights after Edge Removals on LeetCode?

Maximize Sum of Weights after Edge Removals (LeetCode #3367) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Maximize weights while respecting connection limits.

What is the time complexity of Maximize Sum of Weights after Edge Removals?

The optimal solution for Maximize Sum of Weights after Edge Removals runs in O(n log n) time and uses O(n) space. DFS traverses each node once, and sorting edges takes O(n log n).

What is the brute force solution for Maximize Sum of Weights after Edge Removals?

The brute force approach for Maximize Sum of Weights after Edge Removals is Brute Force, with O(n²) time complexity and O(1) space complexity. Try all combinations of edges to find the maximum weight sum while ensuring no node exceeds k edges. This is inefficient but straightforward. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Maximize Sum of Weights after Edge Removals?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar tree problems beforehand.

What are common mistakes when solving Maximize Sum of Weights after Edge Removals?

Forgetting to track visited nodes. Not sorting edges by weight before DFS.

#3367

Maximize Sum of Weights after Edge Removals

Hard

Dynamic ProgrammingTreeDepth-First SearchSortingTree TraversalGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

Use DFS to explore edges and maintain a count of connections per node. Prioritize higher weights while ensuring no node exceeds k connections.

⚙️

Algorithm

3 steps

1Step 1: Build an adjacency list from edges.
2Step 2: Perform DFS from each node, keeping track of the number of edges used.
3Step 3: For each edge, decide to include it based on the weight and current connections.

solution.py18 lines

1def maxWeightOptimal(edges, k):
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    visited = set()
8    def dfs(node, parent):
9        total_weight = 0
10        connections = 0
11        for neighbor, weight in sorted(graph[node], key=lambda x: -x[1]):
12            if neighbor not in visited and connections < k:
13                visited.add(neighbor)
14                total_weight += weight + dfs(neighbor, node)
15                connections += 1
16        return total_weight
17    visited.add(0)
18    return dfs(0, -1)

ℹ

Complexity note: DFS traverses each node once, and sorting edges takes O(n log n).

1Maximize weights while respecting connection limits.
2Use DFS to explore and prioritize edges.

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