What is the time complexity of Shopping Offers?

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

What is the space complexity of Shopping Offers?

The space complexity is O(m^n).

What data structures are used to solve Shopping Offers?

This problem uses Array, Dynamic Programming, Backtracking, Bit Manipulation, Memoization, Bitmask. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Shopping Offers?

Insight 1: Dynamic Programming (Memoization): The problem exhibits overlapping subproblems (e.g., needing to buy the same set of items multiple times during different offer combinations). Therefore, memoization can significantly optimize the solution. Insight 2: Pruning Special Offers: Filtering the special offers to only include those that offer a better deal than buying the items individually is crucial for reducing the search space and improving efficiency.

Which companies ask LeetCode 638. Shopping Offers in interviews?

This problem is frequently asked at Coupang, MathWorks.

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Shopping Offers - LeetCode 638 Solution

Shopping Offers - Complete Solution Guide

Shopping Offers is LeetCode problem 638, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

In LeetCode Store, there are n items to sell. Each item has a price. However, there are some special offers, and a special offer consists of one or more different kinds of items with a sale price. You are given an integer array price where price[i] is the price of the i th item, and an integer array needs where needs[i] is the number of pieces of the i th item you want to buy. You are also given an array special where special[i] is of size n + 1 where special[i][j] is the number of pieces of the

Detailed Explanation

The LeetCode Shopping Offers problem asks us to find the minimum cost to purchase a specific quantity of items given their individual prices and a set of special offers. Each special offer contains a bundle of items at a discounted price. We can use any offer as many times as we want, but we cannot buy more items than needed. The goal is to find the optimal combination of individual purchases and special offers to minimize the total cost.

Solution Approach

The provided solution uses a top-down dynamic programming approach with memoization. It first filters the special offers to include only those that are cheaper than buying the items individually. Then, it recursively explores possible combinations of offers, calculating the cost for each combination. Memoization is used to store the minimum cost for each state (combination of 'needs') to avoid redundant calculations.

Step-by-Step Algorithm

Step 1: **Filter Special Offers:** Iterate through the `special` list. For each offer, calculate the cost of buying the items individually using the `price` list. If the offer's price is less than the individual cost, add it to the `filtered_special` list.
Step 2: **Define Recursive Function `solve(current_needs)`:** This function takes the current `needs` (number of each item required) as input.
Step 3: **Memoization Check:** Before proceeding, check if the result for the current `current_needs` is already stored in the `memo`. If it is, return the stored value.
Step 4: **Calculate Base Cost (Without Offers):** Calculate the cost of buying all the required items individually using the `price` list.
Step 5: **Explore Special Offers:** Iterate through the `filtered_special` list. For each offer:
Step 6: **Check Offer Validity:** Verify that the current `current_needs` can fulfill the offer's requirements (i.e., `current_needs[i] >= offer[i]` for all items).
Step 7: **Apply Offer and Recurs:** If the offer is valid, calculate the `next_needs` by subtracting the offer's items from the `current_needs`. Recursively call `solve(next_needs)` to find the minimum cost to fulfill the remaining needs.
Step 8: **Update Minimum Cost:** Update the `res` (current minimum cost) by taking the minimum of the current `res` and the sum of the offer's price and the cost returned by the recursive call.
Step 9: **Memoize Result:** Store the calculated minimum cost `res` for the current `current_needs` in the `memo`.
Step 10: **Return Result:** Return the calculated minimum cost `res`.

Key Insights

Insight 1: Dynamic Programming (Memoization): The problem exhibits overlapping subproblems (e.g., needing to buy the same set of items multiple times during different offer combinations). Therefore, memoization can significantly optimize the solution.
Insight 2: Pruning Special Offers: Filtering the special offers to only include those that offer a better deal than buying the items individually is crucial for reducing the search space and improving efficiency.
Insight 3: Representing Needs as State: The current 'needs' (the number of each item still required) defines the state of the problem. Each offer we consider transforms the 'needs' into a new state.

Complexity Analysis

Time Complexity: O(n*m^n)

Space Complexity: O(m^n)

Topics

This problem involves: Array, Dynamic Programming, Backtracking, Bit Manipulation, Memoization, Bitmask.

Companies

Asked at: Coupang, MathWorks.

Shopping Offers | LeetCode 638

Medium

Array Dynamic Programming Backtracking Bit Manipulation Memoization Bitmask

Time: O(n*m^n)Space: O(m^n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
        int n = price.size();
        Map<List<Integer>, Integer> memo = new HashMap<>();

        List<List<Integer>> filteredSpecial = new ArrayList<>();
        for (List<Integer> s : special) {
            int costIndividual = 0;
            for (int i = 0; i < n; i++) {
                costIndividual += s.get(i) * price.get(i);
            }
            if (costIndividual > s.get(n)) {
                filteredSpecial.add(s);
            }
        }

        return solve(needs, price, filteredSpecial, memo);
    }

    private int solve(List<Integer> currentNeeds, List<Integer> price, List<List<Integer>> filteredSpecial, Map<List<Integer>, Integer> memo) {
        if (memo.containsKey(currentNeeds)) {
            return memo.get(currentNeeds);
        }

        int n = price.size();
        int res = 0;
        for (int i = 0; i < n; i++) {
            res += currentNeeds.get(i) * price.get(i);
        }

        for (List<Integer> offer : filteredSpecial) {
            boolean validOffer = true;
            for (int i = 0; i < n; i++) {
                if (currentNeeds.get(i) < offer.get(i)) {
                    validOffer = false;
                    break;
                }
            }

            if (validOffer) {
                List<Integer> nextNeeds = new ArrayList<>();
                for (int i = 0; i < n; i++) {
                    nextNeeds.add(currentNeeds.get(i) - offer.get(i));
                }
                res = Math.min(res, offer.get(n) + solve(nextNeeds, price, filteredSpecial, memo));
            }
        }

        memo.put(new ArrayList<>(currentNeeds), res);
        return res;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 638: Shopping Offers - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Backtracking Bit Manipulation Memoization Bitmask

Companies

CoupangMathWorks

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:638

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

10 steps

1
Step 1: **Filter Special Offers:** Iterate through the `special` list. For each offer, calculate the cost of buying the items individually using the `price` list. If the offer's price is less than the individual cost, add it to the `filtered_special` list.
2
Step 2: **Define Recursive Function `solve(current_needs)`:** This function takes the current `needs` (number of each item required) as input.
3
Step 3: **Memoization Check:** Before proceeding, check if the result for the current `current_needs` is already stored in the `memo`. If it is, return the stored value.
4
Step 4: **Calculate Base Cost (Without Offers):** Calculate the cost of buying all the required items individually using the `price` list.
5
Step 5: **Explore Special Offers:** Iterate through the `filtered_special` list. For each offer:
6
Step 6: **Check Offer Validity:** Verify that the current `current_needs` can fulfill the offer's requirements (i.e., `current_needs[i] >= offer[i]` for all items).
7
Step 7: **Apply Offer and Recurs:** If the offer is valid, calculate the `next_needs` by subtracting the offer's items from the `current_needs`. Recursively call `solve(next_needs)` to find the minimum cost to fulfill the remaining needs.
8
Step 8: **Update Minimum Cost:** Update the `res` (current minimum cost) by taking the minimum of the current `res` and the sum of the offer's price and the cost returned by the recursive call.
9
Step 9: **Memoize Result:** Store the calculated minimum cost `res` for the current `current_needs` in the `memo`.
10
Step 10: **Return Result:** Return the calculated minimum cost `res`.

💡

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*m^n)

💾

Space Complexity

O(m^n)

📝

Why?

The time complexity is derived from the fact that each call to `solve` potentially iterates through all special offers. The number of potential `needs` states (arguments for `solve`) is O(m^n) because each of the 'n' items can have up to 'm' different required quantities. The space complexity mirrors the number of potentially stored `needs` states within `memo`.

📖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

Bottom-Up Dynamic Programming

Instead of using recursion and memoization (top-down), a bottom-up dynamic programming approach can be used. Build a table where each entry represents the minimum cost for a specific combination of needs. Iterate through all possible needs combinations and calculate the minimum cost based on individual prices and special offers.

Time: O(n*m^n)Space: O(m^n)

Trade-offs: The bottom-up approach can sometimes be slightly more efficient in terms of constant factors because it avoids the overhead of recursive function calls. However, it may be harder to conceptualize and implement compared to the top-down memoization approach. It also explores the entire space of needs even if some states are unreachable.

✨

Key Insights

1Insight 1: Dynamic Programming (Memoization): The problem exhibits overlapping subproblems (e.g., needing to buy the same set of items multiple times during different offer combinations). Therefore, memoization can significantly optimize the solution.
2Insight 2: Pruning Special Offers: Filtering the special offers to only include those that offer a better deal than buying the items individually is crucial for reducing the search space and improving efficiency.
3Insight 3: Representing Needs as State: The current 'needs' (the number of each item still required) defines the state of the problem. Each offer we consider transforms the 'needs' into a new state.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty special offer list. The code correctly defaults to buying items individually if no special offers are available or are determined to be beneficial.
Edge case 2: When an offer fulfills all needs (all items in needs are satisfied by one or more applications of the offer). The algorithm finds the minimum cost correctly, as it explores different combinations of offers, including the case where only the offer(s) is/are used.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not filtering special offers. Failing to filter special offers that are more expensive than buying the items individually will significantly increase the search space and lead to longer execution times, and potentially Time Limit Exceeded errors.

✗Common mistake 2: Incorrectly implementing the base case or the recursive step. A common error is to not correctly calculate the `next_needs` after applying an offer or to not handle cases where the `needs` become zero.

✅

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: Inventory management and pricing optimization. Companies can use this algorithm to determine the optimal pricing and bundling strategies for their products to maximize profit while minimizing inventory costs.

🏦Real-world scenario 2: Personalized recommendations and discounts. Online retailers can use this algorithm to offer personalized discounts and bundles to customers based on their purchase history and needs, leading to increased sales and customer satisfaction.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

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: September 7, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Shopping Offers"?

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

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