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Solution Approach

Solution Overview

There are 2 main approaches to solve this problem:

  1. Dynamic Programming (Longest Increasing Subsequence Variation) - Complex approach
  2. Binary Indexed Tree (Fenwick Tree) - Complex approach

Approach 1: Dynamic Programming (Longest Increasing Subsequence Variation)

The key insight for solving this problem is to recognize it as a variation of the Longest Increasing Subsequence (LIS) problem, but instead of finding the longest subsequence, we find the subsequence with the maximum sum.

First, we need to sort the players by age (and by score if ages are equal) to ensure that we only consider valid teams. After sorting, we only need to ensure that the scores are non-decreasing as we include players in our team.

Let's define dp[i] as the maximum score we can achieve by considering the first i players and including the i-th player in our team.

For each player i, we look at all previous players j (where j < i). If the score of player j is less than or equal to the score of player i, we can include player i after player j without creating a conflict.

The recurrence relation is:

dp[i] = max(dp[i], dp[j] + scores[i]) for all j < i where scores[j] <= scores[i]

We also need to initialize dp[i] = scores[i] for all i, as we can always form a team with just the i-th player.

The final answer is the maximum value in the dp array.

Algorithm:

  1. Create a list of pairs (age, score) for each player.
  2. Sort the list by age in ascending order (and by score in ascending order if ages are equal).
  3. Initialize a DP array dp of size n, where dp[i] represents the maximum score we can achieve by considering the first i players and including the i-th player.
  4. Initialize dp[i] = scores[i] for all i, as we can always form a team with just the i-th player.
  5. For each player i from 0 to n-1:
  6. a. For each previous player j from 0 to i-1:
  7. i. If scores[j] <= scores[i], update dp[i] = max(dp[i], dp[j] + scores[i]).
  8. Return the maximum value in the dp array.

Time Complexity:

O(n²)

Where n is the number of players. We need to sort the players, which takes O(n log n) time, and then we have two nested loops to fill the DP array, which takes O(n²) time.

Space Complexity:

O(n)

We use a DP array of size n to store the maximum score for each player.

Approach 2: Binary Indexed Tree (Fenwick Tree)

We can optimize the previous approach using a Binary Indexed Tree (BIT) or Fenwick Tree to efficiently query the maximum score for a range of scores.

After sorting the players by age (and by score if ages are equal), we can use a BIT to keep track of the maximum score we can achieve for each possible score value.

For each player, we query the BIT to find the maximum score we can achieve by considering all previous players with scores less than or equal to the current player's score. Then, we update the BIT with the new maximum score for the current player's score.

This approach reduces the time complexity from O(n²) to O(n log n), as each query and update operation on the BIT takes O(log n) time.

Algorithm:

  1. Create a list of pairs (age, score) for each player.
  2. Sort the list by age in ascending order (and by score in ascending order if ages are equal).
  3. Create a BIT of size maxScore + 1, where maxScore is the maximum possible score.
  4. For each player i from 0 to n-1:
  5. a. Query the BIT to find the maximum score maxPrevScore for all scores <= scores[i].
  6. b. Update the BIT with the new maximum score max(current, maxPrevScore + scores[i]) for scores[i].
  7. Return the maximum value in the BIT.

Time Complexity:

O(n log n)

Where n is the number of players. We need to sort the players, which takes O(n log n) time, and then we perform n query and update operations on the BIT, each taking O(log n) time.

Space Complexity:

O(maxScore)

We use a BIT of size maxScore + 1, where maxScore is the maximum possible score.

Pseudocode

solution.pseudo
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function bestTeamScore(scores, ages):
    n = length of scores
    players = []
 
    // Create a list of pairs (age, score)
    for i from 0 to n-1:
        players.append((ages[i], scores[i]))
 
    // Sort by age (and by score if ages are equal)
    sort players by age in ascending order, then by score in ascending order
 
    // Initialize DP array
    dp = new array of size n
    for i from 0 to n-1:
        dp[i] = players[i][1]  // Initialize with the player&apos;s score
 
    // Fill DP array
    for i from 0 to n-1:
        for j from 0 to i-1:
            if players[j][1] <= players[i][1]:  // If score of player j <= score of player i
                dp[i] = max(dp[i], dp[j] + players[i][1])
 
    return max value in dp
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