Divide Players Into Teams of Equal Skill
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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 <= 105skill.lengthis even.1 <= skill[i] <= 1000
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What are the constraints on the size of the `skills` array and the range of values within it?
- Can the `skills` array contain negative numbers or zero?
- If it's impossible to divide the players into teams such that each team has the same skill level, should I return -1 immediately, or should I still attempt to form teams and only return -1 if I'm unable to do so after considering all possible pairs?
- If multiple valid team pairings exist that all result in the same team skill level, does the order in which I pair the players affect the final sum of products, or is the team skill level guaranteed to be unique?
- Is the input array guaranteed to have an even number of elements?
Brute Force Solution
Approach
The brute force method is like trying every single combination to find the answer. We'll explore every possible team pairing to see which ones work according to the problem's rules. This means checking every single way to split up the players.
Here's how the algorithm would work step-by-step:
- Take the first player and pair them up with the second player, calculating the combined skill of this team.
- See if there's another pair of players remaining where the total team skill is the same as the first team's total. If so, you've found a valid team split with two teams.
- If not, try pairing the first player with the third player, then the fourth player, and so on, repeating the check for another valid team.
- If no pair creates a team that enables another valid team, go back and pair the *second* player with the third player, then fourth, and so on. Always look for another pair afterward.
- If you've tried all pairings for the second player, move on to the third and try all possible pairs with the remaining players.
- Continue this process for all players, checking all combinations until you either find a valid division into teams where each team has equal combined skill, or you've exhausted every possibility.
Code Implementation
def divide_players_brute_force(skill):
number_of_players = len(skill)
if number_of_players % 2 != 0:
return -1
for first_player_index in range(number_of_players):
for second_player_index in range(first_player_index + 1, number_of_players):
first_team_skill = skill[first_player_index] + skill[second_player_index]
remaining_players = []
for player_index in range(number_of_players):
if player_index != first_player_index and player_index != second_player_index:
remaining_players.append(skill[player_index])
# Iterate to find a second pair with equal skill
if len(remaining_players) == 2:
if remaining_players[0] + remaining_players[1] == first_team_skill:
return first_team_skill
elif len(remaining_players) > 2:
for third_player_index in range(len(remaining_players)):
for fourth_player_index in range(third_player_index + 1, len(remaining_players)):
if remaining_players[third_player_index] + remaining_players[fourth_player_index] == first_team_skill:
return first_team_skill
# No valid division found
return -1Big(O) Analysis
Time Complexity
O(n^4) – The brute force approach iterates through all possible pairs of players to form the first team. This requires checking approximately n * (n-1) / 2 possible combinations, which is O(n^2). For each of these initial team pairings, the algorithm then searches for another team of equal skill among the remaining players. This secondary search also requires checking pairs among the remaining players, leading to another O(n^2) operation in the worst case. Therefore, the overall time complexity is O(n^2) * O(n^2), resulting in O(n^4).
Space Complexity
O(1) – The brute force approach, as described, primarily involves iterating through different pairings of players without creating significant auxiliary data structures. The algorithm doesn't create temporary lists, hash maps, or recursion stacks that scale with the input size N (the number of players). It primarily utilizes a few variables for indexing to track which players are being paired and to store the combined skill of the current team. Thus, the space used remains constant regardless of the number of players, leading to a space complexity of O(1).
Optimal Solution
Approach
The key idea is to pair the players with the lowest and highest skill levels together. This ensures that each team has a balanced combined skill. Sorting the players beforehand allows us to efficiently find these pairs.
Here's how the algorithm would work step-by-step:
- First, arrange the players in order from the least skilled to the most skilled.
- Then, take the least skilled player and pair them with the most skilled player to form a team.
- Multiply the skills of the two players and record the result.
- Remove these two players from the list.
- Repeat the process of pairing the least and most skilled remaining players until all players are in teams.
- The recorded skill products from each team represent the final answer.
Code Implementation
def divide_players_into_teams(player_skills):
player_skills.sort()
team_skill_products = []
left_index = 0
right_index = len(player_skills) - 1
# Iterate until all players are paired
while left_index < right_index:
# Pair the least skilled with the most skilled player
team_skill_products.append(player_skills[left_index] * player_skills[right_index])
# Move indices towards the center
left_index += 1
right_index -= 1
# All players must result in the same skill level
first_product = team_skill_products[0]
for product in team_skill_products:
if product != first_product:
return -1
# Return the sum of the product of skills for all teams
return sum(team_skill_products)Big(O) Analysis
Time Complexity
O(n log n) – The dominant operation in this algorithm is sorting the players' skills initially. Sorting algorithms like merge sort or quicksort have a time complexity of O(n log n), where n is the number of players. After sorting, pairing the players involves iterating through the sorted list, which takes O(n) time. Since O(n log n) grows faster than O(n) as n increases, the overall time complexity is determined by the sorting step. Therefore, the algorithm's time complexity is O(n log n).
Space Complexity
O(1) – The algorithm sorts the input in place, as described in step 1. While sorting algorithms themselves can have varying space complexities, the explanation assumes in-place sorting. Steps 2-6 operate on the sorted input directly without creating any auxiliary data structures of significant size. Therefore, the auxiliary space used is constant, regardless of the input size N (the number of players).
Edge Cases
| Case | How to Handle |
|---|---|
| Null input array | Throw an IllegalArgumentException or return -1 to indicate invalid input. |
| Array with odd number of elements | Return -1 as an odd number of players cannot be divided into teams of two. |
| Array with only two elements where the sum is not equal to the sum of other pairs. | Check if skills[0] + skills[1] maintains a consistent team skill. |
| Array with elements resulting in integer overflow when multiplied | Use long data type to store the product to prevent overflow. |
| Array with elements that sum to different team skills | Return -1 if there is a mismatch in team skills. |
| Large input array | Ensure the sorting algorithm used has O(n log n) time complexity or the hash map approach has O(n) complexity. |
| Array with duplicate pairs that result in same product, the sum of product must not be counting same pairs more than once. | Hash map or sorting helps to ensure only possible team combinations are calculated and counted only once. |
| Array with negative skill values | The algorithm should correctly handle negative skill values when computing the team skill and product of skill levels. |
Null input array
How to Handle:
Throw an IllegalArgumentException or return -1 to indicate invalid input.
Array with odd number of elements
How to Handle:
Return -1 as an odd number of players cannot be divided into teams of two.
Array with only two elements where the sum is not equal to the sum of other pairs.
How to Handle:
Check if skills[0] + skills[1] maintains a consistent team skill.
Array with elements resulting in integer overflow when multiplied
How to Handle:
Use long data type to store the product to prevent overflow.
Array with elements that sum to different team skills
How to Handle:
Return -1 if there is a mismatch in team skills.
Large input array
How to Handle:
Ensure the sorting algorithm used has O(n log n) time complexity or the hash map approach has O(n) complexity.
Array with duplicate pairs that result in same product, the sum of product must not be counting same pairs more than once.
How to Handle:
Hash map or sorting helps to ensure only possible team combinations are calculated and counted only once.
Array with negative skill values
How to Handle:
The algorithm should correctly handle negative skill values when computing the team skill and product of skill levels.
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