Final Prices With a Special Discount in a Shop - Problem

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.

Input & Output

Example 1 — Basic Discount Case

$ Input: prices = [8,4,6,2,3]

› Output: [4,2,4,2,3]

💡 Note: For item 8: next smaller/equal is 4, so 8-4=4. For item 4: next smaller/equal is 2, so 4-2=2. For item 6: next smaller/equal is 2, so 6-2=4. For item 2: no smaller/equal ahead, so 2-0=2. For item 3: no items ahead, so 3-0=3.

Example 2 — No Discounts Available

$ Input: prices = [1,2,3,4,5]

› Output: [1,2,3,4,5]

💡 Note: Array is strictly increasing, so no item has a smaller or equal item after it. All items keep their original prices.

Example 3 — All Items Get Maximum Discount

$ Input: prices = [10,1,1,6]

› Output: [9,1,1,6]

💡 Note: For item 10: next smaller/equal is 1, so 10-1=9. For item 1 at index 1: next equal is 1 at index 2, so 1-1=0. For item 1 at index 2: no items ahead, so 1-0=1. For item 6: no items ahead, so 6-0=6.

Constraints

1 ≤ prices.length ≤ 500
1 ≤ prices[i] ≤ 1000

Visualization

Tap to expand

Asked in

a Amazon 15 M Microsoft 12

The problem asks us to find final prices after applying special discounts. The key insight is that we need the first item to the right that has a price less than or equal to the current item. The optimal approach uses a monotonic stack processing from right to left. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Check All Future Items

⏱️ Time: O(n²) Space: O(1)

For each item at index i, iterate through all items from i+1 onwards to find the first item with price less than or equal to prices[i]. Subtract this discount from the original price.

Monotonic Stack - Optimal Solution

⏱️ Time: O(n) Space: O(n)

Process items from right to left, maintaining a stack of increasing prices. For each item, pop smaller items from stack (they can't provide discounts), then the stack top is the discount source.

Brute Force - Check All Future Items — Algorithm Steps

Iterate through each item in the array
For each item, scan all future items to find first cheaper one
Apply discount if found, otherwise keep original price

Visualization

Tap to expand

Step-by-Step Walkthrough

Current Item

Start with item at index i

Scan Forward

Check all items from i+1 onwards

Apply Discount

Subtract first cheaper item found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int* solution(int* prices, int n, int* returnSize) {
    *returnSize = n;
    int* result = (int*)malloc(n * sizeof(int));
    
    for (int i = 0; i < n; i++) {
        int discount = 0;
        // Look for first item with price <= prices[i]
        for (int j = i + 1; j < n; j++) {
            if (prices[j] <= prices[i]) {
                discount = prices[j];
                break;
            }
        }
        result[i] = prices[i] - discount;
    }
    
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int prices[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        prices[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int returnSize;
    int* result = solution(prices, n, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops: outer loop runs n times, inner loop runs up to n times for each iteration

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables, result array doesn't count as extra space

✓ Linear Space

53.7K Views

Medium Frequency

~15 min Avg. Time

1.5K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Final Prices With a Special Discount in a Shop - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Future Items — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler