What is the time complexity of Final Prices With a Special Discount in a Shop?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Final Prices With a Special Discount in a Shop?

This problem uses Array, Stack, Monotonic Stack. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Final Prices With a Special Discount in a Shop?

Insight 1: The discount for an item depends on future items' prices. This suggests a need to iterate through the prices array and look ahead for a suitable discount. Insight 2: A brute-force approach involving nested loops is straightforward but might not be the most efficient. A more optimized approach might involve using a stack or other data structure to reduce redundant comparisons.

Which companies ask LeetCode 1475. Final Prices With a Special Discount in a Shop in interviews?

This problem is frequently asked at Dream11.

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Final Prices With a Special Discount in a Shop - LeetCode 1475 Solution

Q: How to solve Final Prices With a Special Discount in a Shop?

The provided Python code uses a brute-force approach. It iterates through the `prices` array. For each item, it iterates through the remaining items to find the first item with a price less than or equal to the current item's price. This price is then used as the discount, and the final price is calculated. If no such item is found, the discount remains 0.

Final Prices With a Special Discount in a Shop - Complete Solution Guide

Final Prices With a Special Discount in a Shop is LeetCode problem 1475, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3.

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.

Detailed Explanation

The problem asks you to calculate the final price of each item in a shop after applying a special discount. The discount for an item is the price of the next item (with a higher index) that has a price less than or equal to the current item's price. If no such item exists, there's no discount. The input is an array `prices` where `prices[i]` represents the price of the i-th item. The output is an array `answer` where `answer[i]` represents the final price of the i-th item after applying the discount.

Solution Approach

The provided Python code uses a brute-force approach. It iterates through the `prices` array. For each item, it iterates through the remaining items to find the first item with a price less than or equal to the current item's price. This price is then used as the discount, and the final price is calculated. If no such item is found, the discount remains 0.

Step-by-Step Algorithm

Step 1: Initialize an empty list `result` to store the final prices.
Step 2: Iterate through the `prices` array using a `for` loop (outer loop).
Step 3: For each item `prices[i]`, initialize a `discount` variable to 0.
Step 4: Iterate through the remaining items starting from `i + 1` (inner loop).
Step 5: If an item `prices[j]` is found such that `prices[j] <= prices[i]`, set `discount` to `prices[j]` and break the inner loop.
Step 6: Calculate the final price as `prices[i] - discount` and append it to the `result` list.
Step 7: After the outer loop completes, return the `result` list.

Key Insights

Insight 1: The discount for an item depends on future items' prices. This suggests a need to iterate through the prices array and look ahead for a suitable discount.
Insight 2: A brute-force approach involving nested loops is straightforward but might not be the most efficient. A more optimized approach might involve using a stack or other data structure to reduce redundant comparisons.
Insight 3: Handling the case where no discount is applicable for an item (no item with a lower price is found later) requires careful attention in the algorithm.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Stack, Monotonic Stack.

Companies

Asked at: Dream11.

Final Prices With a Special Discount in a Shop | LeetCode 1475

Easy

Array Stack Monotonic Stack

Time: O(n^2)Space: O(n)

Solution Code

Available in:

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Problem Info

Description

Problem 1475: Final Prices With a Special Discount in a Shop - Detailed explanation and solution approaches.

Topics

Array Stack Monotonic Stack

Companies

Dream11

Stats

Difficulty:Easy

Languages:2

Companies:1

Problem #:1475

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize an empty list `result` to store the final prices.
2
Step 2: Iterate through the `prices` array using a `for` loop (outer loop).
3
Step 3: For each item `prices[i]`, initialize a `discount` variable to 0.
4
Step 4: Iterate through the remaining items starting from `i + 1` (inner loop).
5
Step 5: If an item `prices[j]` is found such that `prices[j] <= prices[i]`, set `discount` to `prices[j]` and break the inner loop.
6
Step 6: Calculate the final price as `prices[i] - discount` and append it to the `result` list.
7
Step 7: After the outer loop completes, return the `result` list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n^2) because of the nested loops. The outer loop iterates n times, and in the worst case, the inner loop also iterates n times. The space complexity is O(n) because we use a list `result` to store the final prices, which has the same size as the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Monotonic Stack

Use a monotonic decreasing stack. Iterate through the prices. If the current price is less than or equal to the top of the stack, pop elements from the stack until this condition is no longer true, applying discounts based on popped values. Then, push the current price onto the stack.

Time: O(n)Space: O(n)

Trade-offs: This approach significantly improves the time complexity to O(n) by avoiding nested loops, making it more efficient for large input arrays. The space complexity remains O(n) in the worst case (stack size).

✨

Key Insights

1Insight 1: The discount for an item depends on future items' prices. This suggests a need to iterate through the prices array and look ahead for a suitable discount.
2Insight 2: A brute-force approach involving nested loops is straightforward but might not be the most efficient. A more optimized approach might involve using a stack or other data structure to reduce redundant comparisons.
3Insight 3: Handling the case where no discount is applicable for an item (no item with a lower price is found later) requires careful attention in the algorithm.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: Empty input array. The code handles this implicitly by the empty range in the loops - it correctly returns an empty list.
Edge case 2: No discount for any item. The code correctly handles this, as the inner loop will not find a lower price in the remaining items, leaving discount as 0, and returning the original prices.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the discount logic. Beginners might make mistakes in the conditional statements within the inner loop, leading to wrong discount calculations.

✗Common mistake 2: Forgetting to handle the case where no discount is applicable. The code needs to ensure that if the inner loop doesn't find a suitable discount, it doesn't cause errors (the provided solution implicitly does this by initializing discount to 0).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This algorithm can be used in e-commerce applications to calculate final prices after applying discounts based on subsequent purchases or promotions.

🏦Real-world scenario 2: This approach could be used in financial modeling to determine the final value of assets after applying discounts or adjustments based on future events.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Final Prices With a Special Discount in a Shop"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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