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Problem solution · Java

Maximize Sum of Weights after Edge Removals

Maximize Sum of Weights after Edge Removals: a Java solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
50 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Breadth-first search

For Maximize Sum of Weights after Edge Removals, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 50 lines of Java from the credited upstream file 3367.java.
  • The implementation visibly relies on sequence storage, work queue.
  • 3 loop blocks detected, together with recursive traversal.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximize Sum of Weights after Edge Removals · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public long maximizeSumOfWeights(int[][] edges, int k) {    final int n = edges.length + 1;    List<Pair<Integer, Integer>>[] graph = new List[n];    Arrays.setAll(graph, i -> new ArrayList<>());     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      final int w = edge[2];      graph[u].add(new Pair<>(v, w));      graph[v].add(new Pair<>(u, w));    }     return dfs(graph, 0, -1, k).getValue();  }   // Returns  // (the weight sum of the subtree rooted at u with at most k - 1 children,  //  the weight sum of the subtree rooted at u with at most k children).  private Pair<Long, Long> dfs(List<Pair<Integer, Integer>>[] graph, int u, int prev, int k) {    long weightSum = 0;    Queue<Long> diffs = new PriorityQueue<>(Collections.reverseOrder());     for (Pair<Integer, Integer> pair : graph[u]) {      final int v = pair.getKey();      final int w = pair.getValue();      if (v == prev)        continue;      Pair<Long, Long> subResult = dfs(graph, v, u, k);      final long subK1 = subResult.getKey();      final long subK = subResult.getValue();      weightSum += subK;      // If picking (u, v) makes the sum larger, we should pick it.      diffs.offer(Math.max(0L, subK1 - subK + w));    }     long topK1 = 0;    long topK = 0;     for (int i = 0; i < k && !diffs.isEmpty(); ++i) {      if (i < k - 1)        topK1 += diffs.peek();      topK += diffs.poll();    }     return new Pair<>(weightSum + topK1, weightSum + topK);  }}

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