Package g2501_2600.s2538_difference_between_maximum_and_minimum_price_sum

Class Solution

  • All Implemented Interfaces:
    
public final class Solution

    2538 - Difference Between Maximum and Minimum Price Sum.

    Hard

    There exists an undirected and initially unrooted tree with n nodes indexed from 0 to n - 1. You are given the integer n and 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.

    Each node has an associated price. You are given an integer array price, where price[i] is the price of the <code>i<sup>th</sup></code> node.

    The price sum of a given path is the sum of the prices of all nodes lying on that path.

    The tree can be rooted at any node root of your choice. The incurred cost after choosing root is the difference between the maximum and minimum price sum amongst all paths starting at root.

    Return the maximum possible cost amongst all possible root choices.

    Example 1:

    Input: n = 6, edges = [0,1,1,2,1,3,3,4,3,5], price = 9,8,7,6,10,5

    Output: 24

    Explanation: The diagram above denotes the tree after rooting it at node 2. The first part (colored in red) shows the path with the maximum price sum. The second part (colored in blue) shows the path with the minimum price sum.

    • The first path contains nodes 2,1,3,4: the prices are 7,8,6,10, and the sum of the prices is 31.

    • The second path contains the node 2 with the price 7.

    The difference between the maximum and minimum price sum is 24. It can be proved that 24 is the maximum cost.

    Example 2:

    Input: n = 3, edges = [0,1,1,2], price = 1,1,1

    Output: 2

    Explanation: The diagram above denotes the tree after rooting it at node 0. The first part (colored in red) shows the path with the maximum price sum. The second part (colored in blue) shows the path with the minimum price sum.

    • The first path contains nodes 0,1,2: the prices are 1,1,1, and the sum of the prices is 3.

    • The second path contains node 0 with a price 1.

    The difference between the maximum and minimum price sum is 2. It can be proved that 2 is the maximum cost.

    Constraints:

    • <code>1 <= n <= 10<sup>5</sup></code>

    • edges.length == n - 1

    • <code>0 <= a<sub>i</sub>, b<sub>i</sub><= n - 1</code>

    • edges represents a valid tree.

    • price.length == n

    • <code>1 <= pricei<= 10<sup>5</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 Long maxOutput(Integer n, Array<IntArray> edges, IntArray price)

      • Methods inherited from class java.lang.Object

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