#2491

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Divide Players Into Teams of Equal Skill

Divide players into equal skill teams and return the sum of their chemistry, or -1 if impossible.

Problem Statement

You are given an array of even length, where each element represents the skill level of a player. Your task is to divide the players into pairs such that the total skill level of each pair is equal across all teams.

Each team's chemistry is calculated as the product of the skills of its two players. Return the sum of the chemistries of all the teams, or return -1 if no valid division into equal-skill teams is possible.

Examples

Example 1

Input: skill = [3,2,5,1,3,4]

Output: 22

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

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

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 Approach

Sorting and Pairing

Sort the skill array and then use a two-pointer technique to pair the smallest and largest skill values together. This ensures that the total skill of each pair is balanced across teams.

Check Equal Skill Teams

After pairing the players, check if the sum of the skills in each team is equal. If all teams have the same skill sum, proceed to calculate the chemistry of each team. If any pair does not match, return -1.

Efficient Calculation

Calculate the chemistry of each team by multiplying the skill values of the two players in each pair. Sum the chemistries for all valid pairs to get the final result.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(n)

The time complexity of this solution is O(n log n) due to the sorting step, and the space complexity is O(n) to store the skill values and intermediate pairings.

What Interviewers Usually Probe

The candidate correctly identifies that sorting is key to pairing players for equal skill teams.
The candidate applies a two-pointer technique effectively to match players into teams.
The candidate should mention checking for equal skill sums in all pairs before calculating chemistries.

Common Pitfalls or Variants

Common pitfalls

Forgetting to sort the skill array before pairing, which would result in invalid teams.
Not checking if all teams have the same total skill level before calculating chemistry.
Mistaking the chemistry calculation (multiplying skills) as just adding the skills.

Follow-up variants

What if there are constraints on the maximum team size?
How would the solution change if skill[i] could be zero or negative?
What if the problem required calculating chemistry for an odd number of players?

FAQ

How do I divide the players into teams with equal skill levels?

You should sort the skill array and then use a two-pointer technique to pair the smallest and largest values. Check if the sum of each team is equal before calculating the chemistry.

What happens if it's impossible to form equal skill teams?

If no valid way to divide players into equal skill teams exists, return -1.

What is the most efficient way to solve the Divide Players Into Teams of Equal Skill problem?

The most efficient solution involves sorting the array and using a two-pointer approach to ensure teams have equal skill levels.

Why is sorting important for this problem?

Sorting ensures that you can pair players with complementary skill levels, making it easier to balance the total skill of each team.

What if the chemistry of a team isn't calculated correctly?

The chemistry is calculated by multiplying the skills of the two players in the pair. Ensure you are multiplying, not adding the skills.

terminal

Solution

Solution 1: Sorting

To make all 2-person teams have equal skill points, the minimum value must match the maximum value. Therefore, we sort the `skill` array, and then use two pointers $i$ and $j$ to point to the beginning and end of the array respectively, match them in pairs, and judge whether their sum is the same number.

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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

Solution 2: Counting

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the `skill` array.

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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

Continue Practicing

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