# LeetCode 1475. Final Prices With a Special Discount in a Shop

## Description

https://leetcode.com/problems/final-prices-with-a-special-discount-in-a-shop/

Given the 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 array where the `ith` element 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=8 you will receive a discount equivalent to prices=4, therefore, the final price you will pay is 8 - 4 = 4.
For item 1 with price=4 you will receive a discount equivalent to prices=2, therefore, the final price you will pay is 4 - 2 = 2.
For item 2 with price=6 you will receive a discount equivalent to prices=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] <= 10^3`

## Explanation

Iterate the list to find if there is a discount base on finding j, the minimum index such that j > i and prices[j] <= prices[i]

## Python Solution

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

results = []

for i in range(len(prices)):
is_discount = False
for j in range(i + 1, len(prices)):

if prices[i] >= prices[j]:
results.append(prices[i] - prices[j])
is_discount = True
break

if not is_discount:
results.append(prices[i])

return results
``````
• Time Complexity: O(N).
• Space Complexity: O(N).