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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

Solution 1: Monotonic Stack

The problem is essentially to find the first element on the right side that is smaller than each element. We can use a monotonic stack to solve this.

We traverse the array $\textit{prices}$ in reverse order, using the monotonic stack to find the nearest smaller element on the left side of the current element, and then calculate the discount.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{prices}$.

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class Solution:
    def finalPrices(self, prices: List[int]) -> List[int]:
        stk = []
        for i in reversed(range(len(prices))):
            x = prices[i]
            while stk and x < stk[-1]:
                stk.pop()
            if stk:
                prices[i] -= stk[-1]
            stk.append(x)
        return prices
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class Solution {
    public int[] finalPrices(int[] prices) {
        int n = prices.length;
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = n - 1; i >= 0; --i) {
            int x = prices[i];
            while (!stk.isEmpty() && stk.peek() > x) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                prices[i] -= stk.peek();
            }
            stk.push(x);
        }
        return prices;
    }
}
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class Solution {
public:
    vector<int> finalPrices(vector<int>& prices) {
        stack<int> stk;
        for (int i = prices.size() - 1; ~i; --i) {
            int x = prices[i];
            while (!stk.empty() && stk.top() > x) {
                stk.pop();
            }
            if (!stk.empty()) {
                prices[i] -= stk.top();
            }
            stk.push(x);
        }
        return prices;
    }
};
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func finalPrices(prices []int) []int {
    stk := []int{}
    for i := len(prices) - 1; i >= 0; i-- {
        x := prices[i]
        for len(stk) > 0 && stk[len(stk)-1] > x {
            stk = stk[:len(stk)-1]
        }
        if len(stk) > 0 {
            prices[i] -= stk[len(stk)-1]
        }
        stk = append(stk, x)
    }
    return prices
}
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function finalPrices(prices: number[]): number[] {
    const stk: number[] = [];
    for (let i = prices.length - 1; ~i; --i) {
        const x = prices[i];
        while (stk