# Solving Leetcode Interviews in Seconds with AI: Maximize Value of Function in a Ball Passing Game


	# 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](https://www.chatmagic.app), 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 *2<sup>i</sup>* 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
	```python
	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
	```