Introduction

In this blog post, we will explore how to solve the LeetCode problem "1561" 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

There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows: In each step, you will choose any 3 piles of coins (not necessarily consecutive). Of your choice, Alice will pick the pile with the maximum number of coins. You will pick the next pile with the maximum number of coins. Your friend Bob will pick the last pile. Repeat until there are no more piles of coins. Given an array of integers piles where piles[i] is the number of coins in the ith pile. Return the maximum number of coins that you can have. Example 1: Input: piles = [2,4,1,2,7,8] Output: 9 Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one. Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one. The maximum number of coins which you can have are: 7 + 2 = 9. On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal. Example 2: Input: piles = [2,4,5] Output: 4 Example 3: Input: piles = [9,8,7,6,5,1,2,3,4] Output: 18 Constraints: 3 <= piles.length <= 105 piles.length % 3 == 0 1 <= piles[i] <= 104

Explanation

Here's the solution to the problem:

• Sort the piles: Sorting allows us to easily pick the largest and second-largest piles in each round.

• Iterate and accumulate: Starting from the end of the sorted array, pick the second-largest element in each triplet and add it to the total. This is optimal because Alice always takes the largest, and we want the second-largest.

• Runtime & Storage Complexity: O(n log n) time, O(1) space (in-place sort is possible).

Code

    def max_coins(piles: list[int]) -> int:
    """
    Calculates the maximum number of coins you can obtain by picking the second-largest pile in each triplet.

    Args:
        piles: A list of integers representing the number of coins in each pile.

    Returns:
        The maximum number of coins you can have.
    """
    piles.sort()
    n = len(piles)
    coins = 0
    for i in range(n // 3, n, 2):
        coins += piles[i]
    return coins