2944. Minimum Number of Coins for Fruits

Problem Link

Description

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You are given an 1-indexed integer array prices where prices[i] denotes the number of coins needed to purchase the ith fruit.

The fruit market has the following reward for each fruit:

• If you purchase the ith 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] 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 1^st fruit with prices[0] = 3 coins, you are allowed to take the 2^nd fruit for free.
• Purchase the 2^nd fruit with prices[1] = 1 coin, you are allowed to take the 3^rd fruit for free.
• Take the 3^rd fruit for free.

Note that even though you could take the 2^nd fruit for free as a reward of buying 1^st 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 1^st fruit with prices[0] = 1 coin, you are allowed to take the 2^nd fruit for free.
• Take the 2^nd fruit for free.
• Purchase the 3^rd fruit for prices[2] = 1 coin, you are allowed to take the 4^th fruit for free.
• Take the 4^th fruit for free.

Example 3:

Input: prices = [26,18,6,12,49,7,45,45]

Output: 39

Explanation:

• Purchase the 1^st fruit with prices[0] = 26 coin, you are allowed to take the 2^nd fruit for free.
• Take the 2^nd fruit for free.
• Purchase the 3^rd fruit for prices[2] = 6 coin, you are allowed to take the 4^th, 5^th and 6^th (the next three) fruits for free.
• Take the 4^th fruit for free.
• Take the 5^th fruit for free.
• Purchase the 6^th fruit with prices[5] = 7 coin, you are allowed to take the 8^th and 9^th fruit for free.
• Take the 7^th fruit for free.
• Take the 8^th fruit for free.

Note that even though you could take the 6^th fruit for free as a reward of buying 3^rd fruit, you purchase it to receive its reward, which is more optimal.

 

Constraints:

• 1 <= prices.length <= 1000
• 1 <= prices[i] <= 105

Solution

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Python3

class Solution:
    def minimumCoins(self, prices: List[int]) -> int:
        N = len(prices)
        
        @cache
        def go(index, skip):
            if index == N: return 0
            
            res = inf
            
            # useSkip
            if skip > 0:
                res = min(res, go(index + 1, skip - 1))
            
            # take and refresh skip count
            res = min(res, prices[index] + go(index + 1, index + 1))
            
            return res
            
        return go(0, 0)