How do you solve Divide Players Into Teams of Equal Skill on LeetCode?

Divide Players Into Teams of Equal Skill (LeetCode #2491) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Pairing the weakest and strongest players maximizes the chance of equal skill sums.

What is the time complexity of Divide Players Into Teams of Equal Skill?

The optimal solution for Divide Players Into Teams of Equal Skill runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step. The space complexity is O(1) since we are using a constant amount of extra space.

What is the brute force solution for Divide Players Into Teams of Equal Skill?

The brute force approach for Divide Players Into Teams of Equal Skill is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible pairs of players and checking if their total skill can form equal teams. This method is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Divide Players Into Teams of Equal Skill?

Divide Players Into Teams of Equal Skill uses the following concepts: Array, Hash Table, Two Pointers, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Divide Players Into Teams of Equal Skill?

Always consider edge cases, such as the minimum number of players. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice pairing strategies with sorted arrays to build intuition.

What are common mistakes when solving Divide Players Into Teams of Equal Skill?

Not checking if the total skill can be evenly divided into teams. Failing to sort the array before pairing, leading to incorrect team formations.

#2491

Divide Players Into Teams of Equal Skill

Medium

ArrayHash TableTwo PointersSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

The optimal solution leverages sorting and pairing the weakest player with the strongest player. This ensures that we can form teams with equal skill sums efficiently.

⚙️

Algorithm

3 steps

1Step 1: Sort the skill array.
2Step 2: Pair the first element with the last, the second with the second last, and so on.
3Step 3: Calculate the total chemistry based on these pairs.

solution.py14 lines

1# Full working Python code
2
3def divide_players(skill):
4    skill.sort()
5    n = len(skill)
6    target_sum = sum(skill) // (n // 2)
7    chemistry_sum = 0
8
9    for i in range(n // 2):
10        if skill[i] + skill[n - 1 - i] != target_sum:
11            return -1
12        chemistry_sum += skill[i] * skill[n - 1 - i]
13
14    return chemistry_sum

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step. The space complexity is O(1) since we are using a constant amount of extra space.

1Pairing the weakest and strongest players maximizes the chance of equal skill sums.
2Sorting the array allows for efficient pairing and validation of team sums.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.