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Problem solution · Java

Maximum Points After Collecting Coins From All Nodes

Maximum Points After Collecting Coins From All Nodes: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
44 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

Depth-first search

For Maximum Points After Collecting Coins From All Nodes, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 44 lines of Java from the credited upstream file 2920.java.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 Points After Collecting Coins From All Nodes · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maximumPoints(int[][] edges, int[] coins, int k) {    final int n = coins.length;    List<Integer>[] graph = new List[n];    Arrays.setAll(graph, i -> new ArrayList<>());     Integer[][] mem = new Integer[n][MAX_HALVED + 1];     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      graph[u].add(v);      graph[v].add(u);    }     return dfs(graph, 0, /*prev=*/-1, coins, k, /*halved=*/0, mem);  }   private static final int MAX_COIN = 10000;  private static final int MAX_HALVED = (int) (Math.log(MAX_COIN) / Math.log(2)) + 1;   private int dfs(List<Integer>[] graph, int u, int prev, int[] coins, int k, int halved,                  Integer[][] mem) {    // All the children will be 0, so no need to explore.    if (halved > MAX_HALVED)      return 0;    if (mem[u][halved] != null)      return mem[u][halved];     final int val = coins[u] / (1 << halved);    int takeAll = val - k;    int takeHalf = (int) Math.floor(val / 2.0);     for (final int v : graph[u]) {      if (v == prev)        continue;      takeAll += dfs(graph, v, u, coins, k, halved, mem);      takeHalf += dfs(graph, v, u, coins, k, halved + 1, mem);    }     return mem[u][halved] = Math.max(takeAll, takeHalf);  }}

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