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

## Description

You are given an integer 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 integer array answer where answer[i] 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] <= 1000

## Solutions

• class Solution {
public int[] finalPrices(int[] prices) {
int n = prices.length;
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
ans[i] = prices[i];
for (int j = i + 1; j < n; ++j) {
if (prices[j] <= prices[i]) {
ans[i] -= prices[j];
break;
}
}
}
return ans;
}
}

• class Solution {
public:
vector<int> finalPrices(vector<int>& prices) {
int n = prices.size();
vector<int> ans(n);
for (int i = 0; i < n; ++i) {
ans[i] = prices[i];
for (int j = i + 1; j < n; ++j) {
if (prices[j] <= prices[i]) {
ans[i] -= prices[j];
break;
}
}
}
return ans;
}
};

• class Solution:
def finalPrices(self, prices: List[int]) -> List[int]:
ans = []
for i, v in enumerate(prices):
ans.append(v)
for j in range(i + 1, len(prices)):
if prices[j] <= v:
ans[-1] -= prices[j]
break
return ans


• func finalPrices(prices []int) []int {
n := len(prices)
ans := make([]int, n)
for i, v := range prices {
ans[i] = v
for j := i + 1; j < n; j++ {
if prices[j] <= v {
ans[i] -= prices[j]
break
}
}
}
return ans
}

• function finalPrices(prices: number[]): number[] {
const n = prices.length;
const ans = new Array(n);
for (let i = 0; i < n; ++i) {
ans[i] = prices[i];
for (let j = i + 1; j < n; ++j) {
if (prices[j] <= prices[i]) {
ans[i] -= prices[j];
break;
}
}
}
return ans;
}


• /**
* @param {number[]} prices
* @return {number[]}
*/
var finalPrices = function (prices) {
for (let i = 0; i < prices.length; i++) {
for (let j = i + 1; j < prices.length; j++) {
if (prices[i] >= prices[j]) {
prices[i] -= prices[j];
break;
}
}
}
return prices;
};


• class Solution {
/**
* @param Integer[] $prices * @return Integer[] */ function finalPrices($prices) {
for ($i = 0;$i < count($prices);$i++) {
for ($j =$i + 1; $j < count($prices); $j++) { if ($prices[$i] >=$prices[$j]) {$prices[$i] -=$prices[$j]; break; } } } return$prices;
}
}


• impl Solution {
pub fn final_prices(prices: Vec<i32>) -> Vec<i32> {
let n = prices.len();
let mut stack = Vec::new();
let mut res = vec![0; n];
for i in (0..n).rev() {
let price = prices[i];
while !stack.is_empty() && *stack.last().unwrap() > price {
stack.pop();
}
res[i] = price - stack.last().unwrap_or(&0);
stack.push(price);
}
res
}
}