Introduction

In this blog post, we will explore how to solve the LeetCode problem "1475" 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 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

Explanation

Here's the breakdown of the solution:

• Monotonic Stack: Utilize a monotonic stack (decreasing) to efficiently find the next smaller element to the right for each element in the prices array. The stack stores indices.
• Iterate and Calculate: Iterate through the prices array. For each element, maintain the decreasing stack. When encountering a smaller element, pop indices from the stack until the condition is no longer met. Update the answer with the discount.

• Handle No Discount: If the stack is empty after the iteration, no discount applies, and the original price is used.

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

Code

    def final_prices(prices):
    """
    Calculates the final prices after discounts based on the next smaller element.

    Args:
        prices: A list of integers representing the prices of items.

    Returns:
        A list of integers representing the final prices after discounts.
    """

    n = len(prices)
    answer = prices[:]  # Create a copy to store the final prices
    stack = []  # Monotonically decreasing stack (stores indices)

    for i in range(n):
        while stack and prices[stack[-1]] >= prices[i]:
            answer[stack[-1]] -= prices[i]
            stack.pop()
        stack.append(i)

    return answer