2581. Divide Players Into Teams Of Equal Skill¶

Difficulty: Medium

2581. 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 i^th 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¶

class Solution {
    public long dividePlayers(int[] skill) {
        int n = skill.length;
        Arrays.sort(skill);
        int sum = 0, count = 0;
        for (int ele : skill) sum += ele;
        if (sum % (n / 2) != 0) return -1;
        long ans = 0;
        int req = sum / ( n / 2);
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int ele : skill) map.put(ele, map.getOrDefault(ele, 0) + 1);
        for (int i = 0; i < n; i++) {
            int current = skill[i];
            if (map.getOrDefault(current, 0) <= 0) continue;
            int current_req = req - current;
            if (current_req < 0) continue;
            if (map.getOrDefault(current_req , 0) > 0) {
                count++;
                map.put(current_req, map.getOrDefault(current_req , 0) - 1);
                map.put(current, map.getOrDefault(current, 0) - 1);
                ans += (current * current_req);
            }
        }
        if (count == n / 2) return ans;
        return -1;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here