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Best Team With No Conflicts

Difficulty: Medium

Problem Description

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.

Key Insights

  • Players can only be included in the team if there are no conflicts based on age and score.
  • Sorting players by age helps in managing conflicts easily.
  • A dynamic programming approach can be used to maximize the score while ensuring no conflicts occur.

Space and Time Complexity

Time Complexity: O(n^2) where n is the number of players. This is due to the nested iterations used in the dynamic programming solution.
Space Complexity: O(n) for storing the dynamic programming array.

Solution

To solve this problem, we can use a dynamic programming approach. The steps are as follows:

  1. Sort the Players: First, pair each player’s age with their score and sort the players by age. If two players have the same age, sort by score in descending order.

  2. Dynamic Programming Array: Create a DP array where dp[i] represents the maximum score achievable by including players up to index i.

  3. Fill DP Array: For each player, check all previous players to see if they can be included without causing a conflict. If player j (older) has a score less than or equal to player i (younger), update dp[i] to be the maximum of its current value and dp[j] + scores[i].

  4. Result: The result will be the maximum value in the dp array.

Code Solutions

def bestTeamScore(scores, ages):
    players = sorted(zip(ages, scores))
    dp = [0] * len(scores)

    for i in range(len(players)):
        dp[i] = players[i][1]  # Initialize with the score of the player
        for j in range(i):
            if players[j][0] <= players[i][0]:  # No conflict in age
                dp[i] = max(dp[i], dp[j] + players[i][1])  # Update dp[i]

    return max(dp)  # Return the maximum score
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