Package g2901_3000.s2920_maximum_points_after_collecting_coins_from_all_nodes

Class Solution

  • All Implemented Interfaces:
    
public final class Solution

    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 <code>edgesi = a<sub>i</sub>, b<sub>i</sub></code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> 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 <code>node<sub>i</sub></code> 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 <code>node<sub>j</sub></code> present in the subtree of <code>node<sub>i</sub></code>, 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

    • <code>2 <= n <= 10<sup>5</sup></code>

    • <code>0 <= coinsi<= 10<sup>4</sup></code>

    • edges.length == n - 1

    • 0 &lt;= edges[i][0], edges[i][1] &lt; n

    • <code>0 <= k <= 10<sup>4</sup></code>

    • Nested Class Summary

      Nested Classes 
      Modifier and Type Class Description

    • Field Summary

      Fields 
      Modifier and Type Field Description

    • Constructor Summary

      Constructors 
      Constructor Description
      Solution()

    • Enum Constant Summary

      Enum Constants 
      Enum Constant Description

    • Method Summary

      Modifier and Type Method Description
      final Integer maximumPoints(Array<IntArray> edges, IntArray coins, Integer k)

      • Methods inherited from class java.lang.Object

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