LeetCode #1475 — EASY

Final Prices With a Special Discount in a Shop

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given an integer array prices where prices[i] is the price of the i^th item in a shop.

There is a special discount for items in the shop. If you buy the i^th 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 i^th 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Stack

Example 1

[8,4,6,2,3]

Example 2

[1,2,3,4,5]

Example 3

[10,1,1,6]

Step 02

Core Insight

What unlocks the optimal approach

Use brute force: For the ith item in the shop with a loop find the first position j satisfying the conditions and apply the discount, otherwise, the discount is 0.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1475: Final Prices With a Special Discount in a Shop
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;
    }
}

// Accepted solution for LeetCode #1475: Final Prices With a Special Discount in a Shop
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
}

# Accepted solution for LeetCode #1475: Final Prices With a Special Discount in a Shop
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

// Accepted solution for LeetCode #1475: Final Prices With a Special Discount in a Shop
impl Solution {
    pub fn final_prices(mut prices: Vec<i32>) -> Vec<i32> {
        let mut stk: Vec<i32> = Vec::new();
        for i in (0..prices.len()).rev() {
            let x = prices[i];
            while !stk.is_empty() && x < *stk.last().unwrap() {
                stk.pop();
            }
            if let Some(&top) = stk.last() {
                prices[i] -= top;
            }
            stk.push(x);
        }
        prices
    }
}

// Accepted solution for LeetCode #1475: Final Prices With a Special Discount in a Shop
function finalPrices(prices: number[]): number[] {
    const stk: number[] = [];
    for (let i = prices.length - 1; ~i; --i) {
        const x = prices[i];
        while (stk.length && stk.at(-1)! > x) {
            stk.pop();
        }
        prices[i] -= stk.at(-1) || 0;
        stk.push(x);
    }
    return prices;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.