划分技能点相等的团队

You are given a positive integer array skill of even length n where skill[i] denotes the skill of the ith player. Divide the players into n / 2 teams of size 2 such that the total skill of each team is equal.

The chemistry of a team is equal to the product of the skills of the players on that team.

Return the sum of the chemistry of all the teams, or return -1 if there is no way to divide the players into teams such that the total skill of each team is equal.

 

Example 1:

Input: skill = [3,2,5,1,3,4]
Output: 22
Explanation: 
Divide the players into the following teams: (1, 5), (2, 4), (3, 3), where each team has a total skill of 6.
The sum of the chemistry of all the teams is: 1 * 5 + 2 * 4 + 3 * 3 = 5 + 8 + 9 = 22.

Example 2:

Input: skill = [3,4]
Output: 12
Explanation: 
The two players form a team with a total skill of 7.
The chemistry of the team is 3 * 4 = 12.

Example 3:

Input: skill = [1,1,2,3]
Output: -1
Explanation: 
There is no way to divide the players into teams such that the total skill of each team is equal.

 

Constraints:

• 2 <= skill.length <= 105
• skill.length is even.
• 1 <= skill[i] <= 1000

/*
思路：
1. 使用Stream对技能点数组进行排序。
2. 将排序后的数组划分为两部分，使用IntStream分别从头和尾部配对。
3. 检查每个团队的技能点之和是否相等，如果不相等则返回-1。
4. 如果相等，计算团队的化学反应（技能点乘积）并累加。
*/

import java.util.Arrays;
import java.util.stream.IntStream;

public class Solution {
    public int teamChemistry(int[] skill) {
        Arrays.sort(skill); // 对数组进行排序
        int n = skill.length;
        int sum = skill[0] + skill[n - 1]; // 预期的技能点之和
        int chemistry = IntStream.range(0, n / 2) // 从0到n/2
                .map(i -> {
                    int teamSum = skill[i] + skill[n - 1 - i];
                    if (teamSum != sum) {
                        throw new IllegalArgumentException("不匹配"); // 如果有团队的技能点之和不等于预期值，抛出异常
                    }
                    return skill[i] * skill[n - 1 - i]; // 返回团队的化学反应
                }).sum(); // 累加化学反应
        return chemistry; // 返回所有团队的化学反应之和
    }
}

该题解通过先对技能点数组进行排序，然后尝试将最小的元素和最大的元素配对，第二小的元素与第二大的元素配对，以此类推，形成团队。这种配对方式旨在通过将极端值配对来平衡团队的技能点之和，以期望每个团队的技能点之和都相等。如果在配对过程中发现某一对的技能点之和与之前的不同，则返回-1，表明无法形成要求的团队。如果所有配对的技能点之和都相等，则计算每个团队的化学反应（即技能点的乘积），并计算总和返回。

O(n log n)

O(1)