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Formatted question description: https://leetcode.ca/all/2491.html

2491. Divide Players Into Teams of Equal Skill

  • Difficulty: Medium.
  • Related Topics: .
  • Similar Questions: Minimum Moves to Equal Array Elements, Max Number of K-Sum Pairs.

Problem

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

Solution (Java, C++, Python)

  • class Solution {
    public long dividePlayers(int[] skill) {
        Arrays.sort(skill);
        int n = skill.length;
        int t = skill[0] + skill[n - 1];
        long ans = 0;
        for (int i = 0, j = n - 1; i < j; ++i, --j) {
            if (skill[i] + skill[j] != t) {
                return -1;
            }
            ans += (long) skill[i] * skill[j];
        }
        return ans;
    }
}

  • class Solution {
public:
    long long dividePlayers(vector<int>& skill) {
        sort(skill.begin(), skill.end());
        int n = skill.size();
        int t = skill[0] + skill[n - 1];
        long long ans = 0;
        for (int i = 0, j = n - 1; i < j; ++i, --j) {
            if (skill[i] + skill[j] != t) return -1;
            ans += 1ll * skill[i] * skill[j];
        }
        return ans;
    }
};

  • class Solution:
    def dividePlayers(self, skill: List[int]) -> int:
        skill.sort()
        t = skill[0] + skill[-1]
        i, j = 0, len(skill) - 1
        ans = 0
        while i < j:
            if skill[i] + skill[j] != t:
                return -1
            ans += skill[i] * skill[j]
            i, j = i + 1, j - 1
        return ans


  • func dividePlayers(skill []int) (ans int64) {
	sort.Ints(skill)
	n := len(skill)
	t := skill[0] + skill[n-1]
	for i, j := 0, n-1; i < j; i, j = i+1, j-1 {
		if skill[i]+skill[j] != t {
			return -1
		}
		ans += int64(skill[i] * skill[j])
	}
	return
}

  • function dividePlayers(skill: number[]): number {
    const n = skill.length;
    skill.sort((a, b) => a - b);
    const target = skill[0] + skill[n - 1];
    let ans = 0;
    for (let i = 0; i < n >> 1; i++) {
        if (target !== skill[i] + skill[n - 1 - i]) {
            return -1;
        }
        ans += skill[i] * skill[n - 1 - i];
    }
    return ans;
}


  • var dividePlayers = function (skill) {
    const n = skill.length,
        m = n / 2;
    skill.sort((a, b) => a - b);
    const sum = skill[0] + skill[n - 1];
    let ans = 0;
    for (let i = 0; i < m; i++) {
        const x = skill[i],
            y = skill[n - 1 - i];
        if (x + y != sum) return -1;
        ans += x * y;
    }
    return ans;
};


  • impl Solution {
    pub fn divide_players(mut skill: Vec<i32>) -> i64 {
        let n = skill.len();
        skill.sort();
        let target = skill[0] + skill[n - 1];
        let mut ans = 0;
        for i in 0..n >> 1 {
            if skill[i] + skill[n - 1 - i] != target {
                return -1;
            }
            ans += (skill[i] * skill[n - 1 - i]) as i64;
        }
        ans
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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