1475-final-prices-with-a-special-discount-in-a-shop

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Description

You are given an integer array prices where prices[i] is the price of the ith item in a shop.

There is a special discount for items in the shop. If you buy the ith item, then you will receive a discount equivalent to prices[j] where j is the minimum index such that j > i and prices[j] <= prices[i]. Otherwise, you will not receive any discount at all.

Return an integer array answer where answer[i] is the final price you will pay for the ith item of the shop, considering the special discount.

 

Example 1:

Input: prices = [8,4,6,2,3]
Output: [4,2,4,2,3]
Explanation: 
For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8 - 4 = 4.
For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4 - 2 = 2.
For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6 - 2 = 4.
For items 3 and 4 you will not receive any discount at all.

Example 2:

Input: prices = [1,2,3,4,5]
Output: [1,2,3,4,5]
Explanation: In this case, for all items, you will not receive any discount at all.

Example 3:

Input: prices = [10,1,1,6]
Output: [9,0,1,6]

 

Constraints:

• 1 <= prices.length <= 500
• 1 <= prices[i] <= 1000

Solution(Python)

class Solution:
    def finalPrices(self, prices: List[int]) -> List[int]:
        return self.monotonicstack(prices)

    # Time complexity: O(n^2)
    # spac ecomplexity: O(n)
    def bruteforce(self, prices: List[int]) -> List[int]:
        n = len(prices)
        for i in range(n):
            for j in range(i+1, n):
                if prices[j] <= prices[i]:
                    prices[i] -= prices[j]
                    break
                else:
                    continue
        return prices

    # Time complexity: O(n)
    # spac ecomplexity: O(n)
    def monotonicstack(self, prices: List[int]) -> List[int]:
        n = len(prices)
        stack = []
        for i in range(n): # 10 1 1 6
            while stack and prices[stack[-1]] >= prices[i]: # 
                prices[stack[-1]] -= prices[i] # 9 0 0 0
                stack.pop()
            stack.append(i) # stack -> 1

        return prices