# Leetcode - Final Prices With a Special Discount in a Shop Solution

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`

## Solution in Python

```
class Solution:
def finalPrices(self, prices: List[int]) -> List[int]:
discounted_prices = []
for i in range(len(prices)):
x = prices[i]
for j in prices[i+1:]:
if j <= x:
discounted_prices.append(x-j)
break
else:
discounted_prices.append(x)
return discounted_prices
```