Introduction

In this blog post, we will explore how to solve the LeetCode problem "2836" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array receiver of length n and an integer k. n players are playing a ball-passing game. You choose the starting player, i. The game proceeds as follows: player i passes the ball to player receiver[i], who then passes it to receiver[receiver[i]], and so on, for k passes in total. The game's score is the sum of the indices of the players who touched the ball, including repetitions, i.e. i + receiver[i] + receiver[receiver[i]] + ... + receiver(k)[i]. Return the maximum possible score. Notes: receiver may contain duplicates. receiver[i] may be equal to i. Example 1: Input: receiver = [2,0,1], k = 4 Output: 6 Explanation: Starting with player i = 2 the initial score is 2: Pass Sender Index Receiver Index Score 1 2 1 3 2 1 0 3 3 0 2 5 4 2 1 6 Example 2: Input: receiver = [1,1,1,2,3], k = 3 Output: 10 Explanation: Starting with player i = 4 the initial score is 4: Pass Sender Index Receiver Index Score 1 4 3 7 2 3 2 9 3 2 1 10 Constraints: 1 <= receiver.length == n <= 105 0 <= receiver[i] <= n - 1 1 <= k <= 1010

Explanation

Here's an efficient solution to the ball-passing game problem:

• Dynamic Programming (Memoization): Since k can be very large, we can't simulate the entire passing sequence for each starting player. We use memoization to store the player reached after 2ⁱ passes from any starting player. This precomputation helps us quickly determine the player reached after any number of passes by decomposing k into powers of 2.

• Score Calculation: For each potential starting player, we calculate the total score by simulating the passes. Instead of iterating k times, we use the precomputed information to jump ahead in the passing sequence based on the binary representation of k.

• Maximum Score: We iterate through all possible starting players and keep track of the maximum score encountered.

• Runtime Complexity: O(n log k)

• Storage Complexity: O(n log k)

Code

    def max_score(receiver, k):
    n = len(receiver)
    max_log_k = 0
    temp_k = k
    while temp_k > 0:
        max_log_k += 1
        temp_k //= 2

    dp = [[0] * n for _ in range(max_log_k)]  # dp[i][j] is the player reached after 2^i passes from player j

    for j in range(n):
        dp[0][j] = receiver[j]

    for i in range(1, max_log_k):
        for j in range(n):
            dp[i][j] = dp[i - 1][dp[i - 1][j]]

    max_total_score = 0
    for start_player in range(n):
        current_player = start_player
        total_score = 0
        passes_remaining = k

        for i in range(max_log_k):
            if (passes_remaining >> i) & 1:
                total_score += current_player
                current_player = dp[i][current_player]

        total_score += current_player  # Add the final player's index

        max_total_score = max(max_total_score, total_score)

    return max_total_score