This post is part of the Algorithms Problem Solving series.

## Problem description

This is the Final Prices With a Special Discount in a Shop problem. The description looks like this:

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.*

## Examples

```
Input: prices = [8,4,6,2,3]
Output: [4,2,4,2,3]
Input: prices = [1,2,3,4,5]
Output: [1,2,3,4,5]
Input: prices = [10,1,1,6]
Output: [9,0,1,6]
```

## Solution

The solution idea is to iterate over the `prices`

list and for each price, try to get a discount. If it has, apply the discount. Otherwise, the price stays the same.

I built a `get_discount`

function passing the `prices`

list, the `index`

, and the current `price`

.

To get the discount, we need to iterate through the list from the current index + 1 to the end of the `prices`

list. If we find the first price that satisfies the rule, just return this price. If we didn't find any other price, just return 0.

Then we just need to apply the discount by subtracting the `discount`

from the current `price`

and add to the new list of prices with discount.

```
def final_prices(prices):
result = []
for index, price in enumerate(prices):
discount = get_discount(prices, index, price)
result.append(price - discount)
return result
def get_discount(prices, index, price):
for other_price in prices[index + 1:]:
if other_price <= price:
return other_price
return 0
```

We could also modify the `prices`

list in-place.

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