Divide Players Into Teams of Equal Skill

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)

class Solution {
  public long dividePlayers(int[] skill) {
    final int n = skill.length;
    int total = 0;
    int[] count = new int[2001];
    for (int s : skill) {
      total += s;
      count[s]++;
    }

    if (total % (n / 2) != 0) {
      return -1;
    }

    // skill of each team
    final int team = total / (n / 2);
    // number of player
    int seen = 0;
    long chemistry = 0;
    for (int i = 1; i < Math.min(team, 2001); i++) {
      if (count[i] != count[team - i]) {
        return -1;
      }

      if (i == team - i) {
        // player with same skill with even count
        if (count[i] % 2 != 0) {
          return -1;
        }
        chemistry += (long) count[i] / 2 * i * (team - i);
        seen += count[i];
      } else {
        chemistry += (long) count[i] * i * (team - i);
        seen += 2 * count[i];
      }

      count[i] = 0;
      count[team - i] = 0;
    }

    if (seen != n) {
      return -1;
    }

    return chemistry;
  }
}

Explain:

nope.

Complexity:

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