Introduction

In this blog post, we will explore how to solve the LeetCode problem "2944" 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 0-indexed integer array prices where prices[i] denotes the number of coins needed to purchase the (i + 1)th fruit. The fruit market has the following reward for each fruit: If you purchase the (i + 1)th fruit at prices[i] coins, you can get any number of the next i fruits for free. Note that even if you can take fruit j for free, you can still purchase it for prices[j - 1] coins to receive its reward. Return the minimum number of coins needed to acquire all the fruits. Example 1: Input: prices = [3,1,2] Output: 4 Explanation: Purchase the 1st fruit with prices[0] = 3 coins, you are allowed to take the 2nd fruit for free. Purchase the 2nd fruit with prices[1] = 1 coin, you are allowed to take the 3rd fruit for free. Take the 3rd fruit for free. Note that even though you could take the 2nd fruit for free as a reward of buying 1st fruit, you purchase it to receive its reward, which is more optimal. Example 2: Input: prices = [1,10,1,1] Output: 2 Explanation: Purchase the 1st fruit with prices[0] = 1 coin, you are allowed to take the 2nd fruit for free. Take the 2nd fruit for free. Purchase the 3rd fruit for prices[2] = 1 coin, you are allowed to take the 4th fruit for free. Take the 4th fruit for free. Example 3: Input: prices = [26,18,6,12,49,7,45,45] Output: 39 Explanation: Purchase the 1st fruit with prices[0] = 26 coin, you are allowed to take the 2nd fruit for free. Take the 2nd fruit for free. Purchase the 3rd fruit for prices[2] = 6 coin, you are allowed to take the 4th, 5th and 6th (the next three) fruits for free. Take the 4th fruit for free. Take the 5th fruit for free. Purchase the 6th fruit with prices[5] = 7 coin, you are allowed to take the 8th and 9th fruit for free. Take the 7th fruit for free. Take the 8th fruit for free. Note that even though you could take the 6th fruit for free as a reward of buying 3rd fruit, you purchase it to receive its reward, which is more optimal. Constraints: 1 <= prices.length <= 1000 1 <= prices[i] <= 105

Explanation

Here's an efficient solution to the problem:

• Dynamic Programming: The core idea is to use dynamic programming to explore the optimal choices at each fruit. dp[i] stores the minimum cost to acquire fruits up to index i.
• Two Choices: For each fruit i, we have two choices: either buy it and get the next i fruits for free or skip it and take it for free if we've bought a previous fruit that allows us to.

• Base Case: dp[0] = prices[0] as we have to buy the first fruit.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    def min_coins(prices):
    n = len(prices)
    dp = [0] * n

    dp[0] = prices[0]

    for i in range(1, n):
        # Option 1: Buy the current fruit
        cost_buy = prices[i]

        # Cost up to the previous fruit + cost of buying the current fruit
        cost_buy += dp[i-1] if i > 0 else 0

        # Option 2: Don't buy the current fruit (take it for free if possible).
        cost_skip = float('inf')
        can_skip = False
        for j in range(i):
            if i <= j + j:  # i <= bought_fruit_index + reward of buying bought_fruit_index-th fruit
                can_skip = True
                break

        if can_skip:
            cost_skip = dp[i - 1] if i > 0 else 0
        else:
            cost_skip = float('inf')

        dp[i] = min(cost_buy, cost_skip if i > 0 else float('inf'))

        #Optimization. If current optimal cost is same as that of i-1-th fruit
        #then it is bound to give the same result. So avoid recalculating next
        #values for them.
        if i > 0 and dp[i] == dp[i-1]:
            prices[i] = float('inf')

    return dp[n - 1]