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Problem solution · Java

Maximum Sum of Edge Values in a Graph

Maximum Sum of Edge Values in a Graph: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
71 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

Sliding window or two pointers

For Maximum Sum of Edge Values in a Graph, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 71 lines of Java from the credited upstream file 3547.java.
  • The implementation visibly relies on sequence storage, work queue.
  • 7 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 codeMaximum Sum of Edge Values in a Graph · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public long maxScore(int n, int[][] edges) {    long ans = 0;    List<Integer>[] graph = new List[n];    List<Integer> cycleSizes = new ArrayList<>(); // components where all nodes have degree 2    List<Integer> pathSizes = new ArrayList<>();  // components that are not cycleSizes    boolean[] seen = new boolean[n];    Arrays.setAll(graph, i -> new ArrayList<>());     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      graph[u].add(v);      graph[v].add(u);    }     for (int i = 0; i < n; ++i) {      if (seen[i])        continue;      List<Integer> component = getComponent(graph, i, seen);      final boolean allDegree2 = component.stream().allMatch(u -> graph[u].size() == 2);      if (allDegree2)        cycleSizes.add(component.size());      else if (component.size() > 1)        pathSizes.add(component.size());    }     for (final int cycleSize : cycleSizes) {      ans += calculateScore(n - cycleSize + 1, n, true);      n -= cycleSize;    }     Collections.sort(pathSizes, Collections.reverseOrder());     for (final int pathSize : pathSizes) {      ans += calculateScore(n - pathSize + 1, n, false);      n -= pathSize;    }     return ans;  }   private List<Integer> getComponent(List<Integer>[] graph, int start, boolean[] seen) {    List<Integer> component = new ArrayList<>(List.of(start));    seen[start] = true;    for (int i = 0; i < component.size(); ++i) {      final int u = component.get(i);      for (final int v : graph[u]) {        if (seen[v])          continue;        component.add(v);        seen[v] = true;      }    }    return component;  }   private long calculateScore(int left, int right, boolean isCycle) {    Deque<Long> window = new ArrayDeque<>();    window.offerLast((long) right);    window.offerLast((long) right);    long score = 0;    for (int value = right - 1; value >= left; --value) {      final long windowValue = window.pollFirst();      score += windowValue * value;      window.offerLast((long) value);    }    return score + window.peekFirst() * window.peekLast() * (isCycle ? 1 : 0);  }}

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