Sorting, Two-Pointers and Long

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This question just requires a sorting (nlogn), two-pointers (n) and conversion to long. Code is down below, cheers, ACC.

Divide Players Into Teams of Equal Skill - LeetCode

2491. Divide Players Into Teams of Equal Skill
Medium

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

public long DividePlayers(int[] skill)
{
    Array.Sort(skill);

    long retVal = (long)skill[0] * skill[skill.Length - 1];
    long sum = (long)skill[0] + skill[skill.Length - 1];

    int left = 1;
    int right = skill.Length - 2;
    while (left < right)
    {
        long currentSum = (long)skill[left] + skill[right];
        if (currentSum != sum) return -1;
        retVal += (long)skill[left] * skill[right];
        left++;
        right--;
    }

    return retVal;
}

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