2491. Divide Players Into Teams of Equal Skill LeetCode Solution

In this guide, you will get 2491. Divide Players Into Teams of Equal Skill LeetCode Solution with the best time and space complexity. The solution to Divide Players Into Teams of Equal Skill problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Divide Players Into Teams of Equal Skill solution in C++
4. Divide Players Into Teams of Equal Skill solution in Java
5. Divide Players Into Teams of Equal Skill solution in Python
6. Additional Resources

Problem Statement of Divide Players Into Teams of Equal Skill

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

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(n)

2491. Divide Players Into Teams of Equal Skill LeetCode Solution in C++

class Solution {
 public:
  long long dividePlayers(vector<int>& skill) {
    const int n = skill.size();
    const int teamSkill = accumulate(skill.begin(), skill.end(), 0) / (n / 2);
    long ans = 0;
    unordered_map<int, int> count;

    for (const int s : skill)
      ++count[s];

    for (const auto& [s, freq] : count) {
      const int requiredSkill = teamSkill - s;
      if (const auto it = count.find(requiredSkill);
          it == count.cend() || it->second != freq)
        return -1;
      ans += static_cast<long>(s) * requiredSkill * freq;
    }

    return ans / 2;
  }
};
/* code provided by PROGIEZ */

2491. Divide Players Into Teams of Equal Skill LeetCode Solution in Java

class Solution {
  public long dividePlayers(int[] skill) {
    final int n = skill.length;
    final int teamSkill = Arrays.stream(skill).sum() / (n / 2);
    long ans = 0;
    Map<Integer, Integer> count = new HashMap<>();

    for (final int s : skill)
      count.merge(s, 1, Integer::sum);

    for (Map.Entry<Integer, Integer> entry : count.entrySet()) {
      final int s = entry.getKey();
      final int freq = entry.getValue();
      final int requiredSkill = teamSkill - s;
      if (count.getOrDefault(requiredSkill, 0) != freq)
        return -1;
      ans += (long) s * requiredSkill * freq;
    }

    return ans / 2;
  }
}
// code provided by PROGIEZ

2491. Divide Players Into Teams of Equal Skill LeetCode Solution in Python

class Solution:
  def dividePlayers(self, skill: list[int]) -> int:
    n = len(skill)
    teamSkill = sum(skill) // (n // 2)
    ans = 0
    count = collections.Counter(skill)

    for s, freq in count.items():
      requiredSkill = teamSkill - s
      if count[requiredSkill] != freq:
        return -1
      ans += s * requiredSkill * freq

    return ans // 2
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

Happy Coding! Keep following PROGIEZ for more updates and solutions.