Best Team With No Conflicts

Medium

Morgan Stanley

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Topics:

ArraysDynamic Programming

You are the manager of a basketball team. For the upcoming tournament, you want to choose the team with the highest overall score. The score of the team is the sum of scores of all the players in the team.

However, the basketball team is not allowed to have conflicts. A conflict exists if a younger player has a strictly higher score than an older player. A conflict does not occur between players of the same age.

Given two lists, scores and ages, where each scores[i] and ages[i] represents the score and age of the i^th player, respectively, return the highest overall score of all possible basketball teams.

Example 1:

Input: scores = [1,3,5,10,15], ages = [1,2,3,4,5]
Output: 34
Explanation: You can choose all the players.

Example 2:

Input: scores = [4,5,6,5], ages = [2,1,2,1]
Output: 16
Explanation: It is best to choose the last 3 players. Notice that you are allowed to choose multiple people of the same age.

Example 3:

Input: scores = [1,2,3,5], ages = [8,9,10,1]
Output: 6
Explanation: It is best to choose the first 3 players.

Constraints:

1 <= scores.length, ages.length <= 1000
scores.length == ages.length
1 <= scores[i] <= 10⁶
1 <= ages[i] <= 1000

Try coding it in LeetCode

Best Team With No Conflicts

Medium

Morgan Stanley

4 views

Topics:

ArraysDynamic Programming

You are the manager of a basketball team. For the upcoming tournament, you want to choose the team with the highest overall score. The score of the team is the sum of scores of all the players in the team.

However, the basketball team is not allowed to have conflicts. A conflict exists if a younger player has a strictly higher score than an older player. A conflict does not occur between players of the same age.

Given two lists, scores and ages, where each scores[i] and ages[i] represents the score and age of the i^th player, respectively, return the highest overall score of all possible basketball teams.

Example 1:

Input: scores = [1,3,5,10,15], ages = [1,2,3,4,5]
Output: 34
Explanation: You can choose all the players.

Example 2:

Input: scores = [4,5,6,5], ages = [2,1,2,1]
Output: 16
Explanation: It is best to choose the last 3 players. Notice that you are allowed to choose multiple people of the same age.

Example 3:

Input: scores = [1,2,3,5], ages = [8,9,10,1]
Output: 6
Explanation: It is best to choose the first 3 players.

Constraints:

1 <= scores.length, ages.length <= 1000
scores.length == ages.length
1 <= scores[i] <= 10⁶
1 <= ages[i] <= 1000

Try coding it in LeetCode

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 `scores` and `ages` arrays?
Can `scores` or `ages` contain negative values?
If there are multiple teams with the same maximum total score, can I return any one of them?
If `ages[i]` is the same for multiple players, is there any specific ordering preference for considering their scores in the team?
If the input arrays are empty or null, what value should I return?

Brute Force Solution

Approach

The brute force strategy for finding the best team with no conflicts involves checking every possible combination of team members. It's like trying out every possible group of people and seeing if they work well together. This means we'll need to evaluate each possible subset of candidates.

Here's how the algorithm would work step-by-step:

First, consider a team with just one person. Check if this person has any conflicts.
Next, consider all possible teams of two people. For each of these teams, check if there are any conflicts between the two team members.
Then, consider all possible teams of three people. Check for conflicts between every pair of people within each of these teams.
Continue this process, increasing the team size by one each time, until you've considered a team with all the people.
For each team you considered, if there are no conflicts, calculate the total score of the team.
Finally, compare the total scores of all the conflict-free teams and choose the team with the highest score. That's your best team.

Code Implementation

def best_team_with_no_conflicts_brute_force(scores, ages):
    number_of_players = len(scores)
    best_team_score = 0

    for i in range(1 << number_of_players):
        current_team_scores = []
        current_team_ages = []
        current_team_score = 0

        for j in range(number_of_players):
            if (i >> j) & 1:
                current_team_scores.append(scores[j])
                current_team_ages.append(ages[j])
                current_team_score += scores[j]

        is_valid_team = True

        # Need to check if the current team is valid
        for first_player_index in range(len(current_team_scores)):
            for second_player_index in range(first_player_index + 1, len(current_team_scores)):
                if current_team_ages[first_player_index] < current_team_ages[second_player_index] and current_team_scores[first_player_index] > current_team_scores[second_player_index]:
                    is_valid_team = False
                    break
                if current_team_ages[first_player_index] > current_team_ages[second_player_index] and current_team_scores[first_player_index] < current_team_scores[second_player_index]:
                    is_valid_team = False
                    break

            if not is_valid_team:
                break

        # Only update the best team if current team has no conflicts and a better score
        if is_valid_team:
            best_team_score = max(best_team_score, current_team_score)

    return best_team_score

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm iterates through all possible subsets of the input of size n, representing the candidates. Generating all subsets requires considering each candidate for inclusion or exclusion, leading to 2^n possible combinations. For each subset, the algorithm checks for conflicts between pairs of individuals within that subset. The conflict checking within each subset does not change the overall complexity since the number of subsets dominates. Therefore, the time complexity is O(2^n).

Space Complexity

O(1) – The brute force approach, as described, primarily iterates through different team combinations and checks for conflicts. It calculates the score of valid teams but does not explicitly create any data structures to store these combinations or intermediate results beyond what's immediately needed for comparison. Thus, the algorithm uses a constant amount of extra space, mainly for storing the current team's score and the best team's score found so far. Therefore, the auxiliary space complexity is O(1), independent of the number of people (N).

Optimal Solution

Approach

To find the best team, we need to consider both skill and age, avoiding conflicts where an older person has a lower skill than a younger person. The key is to sort people by age first, and then use a clever trick to track the best possible team skill without actually trying every team combination.

Here's how the algorithm would work step-by-step:

First, put all the people in order based on their age, from youngest to oldest. If some people are the same age, we can order those within the group in any order we want, it won't affect the correctness.
Now, imagine building the team one person at a time, going from youngest to oldest.
As you consider each person, you have two choices: either include them on the team or leave them out.
To make the right choice, track the best possible team skill you can achieve *up to that point*, if we include the current person as the last person on the team.
When you reach the current person, check every other previous candidate. If that person has more skill than the current person, add the current person's skill to the total skills of the best team ending with that older candidate.
Keep track of the biggest team skill you have seen so far at each step.
The answer is the biggest skill of the team you found at the end of the process.

Code Implementation

def best_team_score(scores, ages):
    people = sorted(zip(ages, scores))
    number_of_people = len(scores)
    dp = [0] * number_of_people
    max_team_score = 0

    for i in range(number_of_people):
        dp[i] = people[i][1]

        # Iterate through previous candidates.
        for j in range(i):
            # Check for conflicts based on skill.
            if people[j][1] <= people[i][1]:
                dp[i] = max(dp[i], dp[j] + people[i][1])

        # Keep track of the maximum team score found so far.
        max_team_score = max(max_team_score, dp[i])

    return max_team_score

Big(O) Analysis

Time Complexity

O(n²) – The described solution involves sorting n people by age, which typically takes O(n log n) time, but this is dominated by the subsequent steps. The core logic iterates through each person (n), and for each person, checks every previous person to determine the best possible team skill (another n). This nested iteration results in roughly n * n comparisons. Thus, the dominant factor is O(n * n) which simplifies to O(n²).

Space Complexity

O(N) – The provided algorithm sorts the input which may or may not be in place. However, the core of the algorithm involves tracking the best possible team skill up to each person when considered as the last member of the team. This implies creating an auxiliary array or list of size N to store the maximum team skill ending with each of the N individuals. Therefore, the auxiliary space required scales linearly with the number of people, N. This results in a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty `scores` or `ages` array	Return 0, as no players can form a team.
`scores` and `ages` arrays have different lengths	Throw an IllegalArgumentException or return an error code, as the input is invalid.
Single player in both `scores` and `ages` arrays	Return the single player's score, as they trivially form a valid team.
All players have the same age	Return the sum of all scores, as there will be no conflicts.
All players have the same score	Find the largest age and select players only up to the largest age.
Large input size causing potential performance issues (e.g., > 10^5 players)	Ensure the algorithm has a time complexity of O(n log n) or better using sorting and dynamic programming for efficient processing.
Integer overflow when summing scores	Use a long data type for accumulating the team score to prevent potential integer overflow.
Players with very large scores and ages	Ensure the solution handles the extreme boundary values of the integer type without causing errors.