Package g2901_3000.s2920_maximum_points_after_collecting_coins_from_all_nodes

Class Solution

java.lang.Object
g2901_3000.s2920_maximum_points_after_collecting_coins_from_all_nodes.Solution
public class Solution extends java.lang.Object
2920 - Maximum Points After Collecting Coins From All Nodes.

Hard

There exists an undirected tree rooted at node 0 with n nodes labeled from 0 to n - 1. You are given a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. You are also given a 0-indexed array coins of size n where coins[i] indicates the number of coins in the vertex i, and an integer k.

Starting from the root, you have to collect all the coins such that the coins at a node can only be collected if the coins of its ancestors have been already collected.

Coins at nodei can be collected in one of the following ways:

  • Collect all the coins, but you will get coins[i] - k points. If coins[i] - k is negative then you will lose abs(coins[i] - k) points.
  • Collect all the coins, but you will get floor(coins[i] / 2) points. If this way is used, then for all the nodej present in the subtree of nodei, coins[j] will get reduced to floor(coins[j] / 2).

Return the maximum points you can get after collecting the coins from all the tree nodes.

Example 1:

Input: edges = [[0,1],[1,2],[2,3]], coins = [10,10,3,3], k = 5

Output: 11

Explanation:

Collect all the coins from node 0 using the first way. Total points = 10 - 5 = 5.

Collect all the coins from node 1 using the first way. Total points = 5 + (10 - 5) = 10.

Collect all the coins from node 2 using the second way so coins left at node 3 will be floor(3 / 2) = 1. Total points = 10 + floor(3 / 2) = 11.

Collect all the coins from node 3 using the second way. Total points = 11 + floor(1 / 2) = 11.

It can be shown that the maximum points we can get after collecting coins from all the nodes is 11.

Example 2:

Input: edges = [[0,1],[0,2]], coins = [8,4,4], k = 0

Output: 16

Explanation: Coins will be collected from all the nodes using the first way. Therefore, total points = (8 - 0) + (4 - 0) + (4 - 0) = 16.

Constraints:

  • n == coins.length
  • 2 <= n <= 105
  • 0 <= coins[i] <= 104
  • edges.length == n - 1
  • 0 <= edges[i][0], edges[i][1] < n
  • 0 <= k <= 104

  • Constructor Summary

    Constructors
    Constructor
    Description
    Solution()
     

  • Method Summary

    Modifier and Type
    Method
    Description
    int
    maximumPoints(int[][] edges, int[] coins, int k)
     

    Methods inherited from class java.lang.Object

    clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • maximumPoints

      public int maximumPoints(int[][] edges, int[] coins, int k)