Introduction

In this blog post, we will explore how to solve the LeetCode problem "2218" 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 n piles of coins on a table. Each pile consists of a positive number of coins of assorted denominations. In one move, you can choose any coin on top of any pile, remove it, and add it to your wallet. Given a list piles, where piles[i] is a list of integers denoting the composition of the ith pile from top to bottom, and a positive integer k, return the maximum total value of coins you can have in your wallet if you choose exactly k coins optimally. Example 1: Input: piles = [[1,100,3],[7,8,9]], k = 2 Output: 101 Explanation: The above diagram shows the different ways we can choose k coins. The maximum total we can obtain is 101. Example 2: Input: piles = [[100],[100],[100],[100],[100],[100],[1,1,1,1,1,1,700]], k = 7 Output: 706 Explanation: The maximum total can be obtained if we choose all coins from the last pile. Constraints: n == piles.length 1 <= n <= 1000 1 <= piles[i][j] <= 105 1 <= k <= sum(piles[i].length) <= 2000

Explanation

Here's the approach and code for solving the coin pile maximization problem:

• Dynamic Programming: We use dynamic programming to store the maximum value achievable for different numbers of coins taken up to a certain pile. dp[i][j] stores the maximum value when considering the first i piles and taking exactly j coins.
• Iterative Calculation: We iterate through the piles and the number of coins to take. For each pile, we consider all possible numbers of coins to take from that pile (including zero).

• Prefix Sum Optimization: To efficiently calculate the value of taking a certain number of coins from a pile, we precompute the prefix sum of each pile.

• Runtime Complexity: O(n*k*k), where n is the number of piles and k is the maximum number of coins to take. Storage Complexity: O(n*k).

Code

    def max_value_of_coins(piles, k):
    n = len(piles)
    dp = [[0] * (k + 1) for _ in range(n + 1)]

    for i in range(1, n + 1):
        pile = piles[i - 1]
        pile_len = len(pile)
        prefix_sum = [0] * (pile_len + 1)
        for j in range(1, pile_len + 1):
            prefix_sum[j] = prefix_sum[j - 1] + pile[j - 1]

        for j in range(k + 1):
            for x in range(min(j, pile_len) + 1):
                dp[i][j] = max(dp[i][j], dp[i - 1][j - x] + prefix_sum[x])

    return dp[n][k]