Program to find Final Prices With a Special Discount in a Shop in Python

Given an array of prices, we need to find the final prices after applying a special discount. For each item at index i, we get a discount equal to the price of the first item at index j where j > i and prices[j] ? prices[i].

Problem Understanding

The discount rule works as follows ?

• For item at index i, look for the first item at index j where j > i

• If prices[j] ? prices[i], then discount = prices[j]

• Final price = prices[i] − prices[j]

• If no such j exists, no discount is applied

Example Walkthrough

For prices = [16, 8, 12, 4, 6] ?

• Item 0 (price=16): First smaller/equal item is at index 1 (price=8), so final price = 16 − 8 = 8

• Item 1 (price=8): First smaller/equal item is at index 3 (price=4), so final price = 8 − 4 = 4

• Item 2 (price=12): First smaller/equal item is at index 3 (price=4), so final price = 12 − 4 = 8

• Item 3 (price=4): No item after index 3 is ? 4, so no discount, final price = 4

• Item 4 (price=6): No items after index 4, so no discount, final price = 6

Solution Implementation

def final_prices_with_discount(prices):
    result = prices.copy()  # Create a copy to avoid modifying original
    
    for i in range(len(prices)):
        for j in range(i + 1, len(prices)):
            if prices[j] <= prices[i]:
                result[i] = prices[i] - prices[j]
                break
    
    return result

# Test the solution
prices = [16, 8, 12, 4, 6]
final_prices = final_prices_with_discount(prices)
print("Original prices:", prices)
print("Final prices:   ", final_prices)

Original prices: [16, 8, 12, 4, 6]
Final prices:    [8, 4, 8, 4, 6]

Step-by-Step Algorithm

The algorithm follows these steps ?

1. Create a copy of the original prices array

2. For each item at index i from 0 to length-1:

   • Search for the first item at index j where j > i

   • If prices[j] ? prices[i], apply discount and break

   • If no such item found, keep original price

3. Return the result array

Alternative Test Case

def final_prices_with_discount(prices):
    result = prices.copy()
    
    for i in range(len(prices)):
        for j in range(i + 1, len(prices)):
            if prices[j] <= prices[i]:
                result[i] = prices[i] - prices[j]
                break
    
    return result

# Test with different prices
test_prices = [10, 1, 1, 6]
print("Test prices:", test_prices)
print("Final prices:", final_prices_with_discount(test_prices))

Test prices: [10, 1, 1, 6]
Final prices: [9, 0, 1, 6]

Time and Space Complexity

Conclusion

This solution uses a brute force approach to find the first eligible discount for each item. The nested loop structure ensures we find the nearest qualifying discount, making it suitable for the given problem constraints.