Final Prices With a Special Discount in a Shop

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: next equal is 1, so 1-1=0. Wait, that's wrong. Let me recalculate: For 1 at index 1: next smaller/equal is 1 at index 2, so 1-1=0. But 0 is wrong. Actually 1-1=0 is correct, but the expected should be [9,0,1,6]. Let me fix: For last 1: no items ahead, so 1-0=1. For 6: no items ahead, so 6-0=6.

Constraints

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

Visualization

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Understanding the Visualization

Input Array

Array of item prices [8,4,6,2,3]

Find Discounts

For each item, find first item to right with price ≤ current

Apply Discounts

Subtract discount from original price to get final price

Key Takeaway

🎯 Key Insight: Use monotonic stack to efficiently find the next smaller/equal element for each position

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

Approach	Time	Space	Notes
✓ Monotonic Stack - Optimal Solution	O(n)	O(n)	Use a stack to efficiently track items waiting for discounts
Brute Force - Check All Future Items	O(n²)	O(1)	For each item, scan all future items to find the first cheaper one

Monotonic Stack - Optimal Solution — Algorithm Steps

Traverse array from right to left
Maintain stack of indices with increasing prices
For each item, find discount from stack top after cleanup

Visualization

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Step-by-Step Walkthrough

Process Right to Left

Start from the last element

Maintain Stack

Keep only elements that can provide discounts

Find Discount

Stack top gives the discount amount

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));
    int* stack = (int*)malloc(n * sizeof(int));
    int stackTop = -1;
    
    // Process from right to left
    for (int i = n - 1; i >= 0; i--) {
        // Remove elements that can't provide discount
        while (stackTop >= 0 && prices[stack[stackTop]] > prices[i]) {
            stackTop--;
        }
        
        // Calculate final price
        if (stackTop >= 0) {
            result[i] = prices[i] - prices[stack[stackTop]];
        } else {
            result[i] = prices[i];  // No discount available
        }
        
        // Add current index to stack
        stack[++stackTop] = i;
    }
    
    free(stack);
    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(n)

Each element is pushed and popped from stack at most once, making it linear time

✓ Linear Growth

Space Complexity

O(n)

Stack can store up to n indices in worst case (strictly increasing prices)

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Monotonic Stack - Optimal Solution — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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