## 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[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] <= 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).