Problem Statement
Shopping Offers
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][0], special[i][1], ..., special[i][n-1] represents the number of pieces of each item and special[i][n] represents the price of the offer.
Return the lowest price you have to pay for exactly certain items as given, where you could make optimal use of the special offers. You are not allowed to buy more items than you want, even if that would lower the overall price. You could use any of the special offers as many times as you want.
Examples
Example 1:
Input: price = [2, 5], special = [[3, 0, 5], [1, 2, 10]], needs = [3, 2]
Output: 14
Explanation: There are two kinds of items, A and B. Their prices are $2 and $5 respectively.
In special offer 1, you can pay $5 for 3A and 0B.
In special offer 2, you can pay $10 for 1A and 2B.
You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.
Example 2:
Input: price = [2, 3, 4], special = [[1, 1, 0, 4], [2, 2, 1, 9]], needs = [1, 2, 1]
Output: 11
Explanation: The price of A is $2, and $3 for B, and $4 for C.
You may pay $4 for 1A and 1B, and $9 for 2A, 2B and 1C.
You need to buy 1A, 2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, and $4 for 1C.
You cannot use special offer #2, since it would make you buy 2A, 2B and 1C, which exceeds your needs.
Example 3:
Input: price = [2, 3, 4], special = [[1, 1, 0, 4], [2, 2, 1, 9]], needs = [0, 0, 0]
Output: 0
Explanation: You don't need to buy anything, so the total price is $0.
Constraints
- n == price.length
- n == needs.length
- 1 <= n <= 6
- 0 <= price[i] <= 10
- 0 <= needs[i] <= 10
- 1 <= special.length <= 100
- special[i].length == n + 1
- 0 <= special[i][j] <= 50
Problem Breakdown
To solve this problem, we need to:
- This problem can be solved using dynamic programming or recursion with memoization.
- For each special offer, we need to decide whether to use it or not, and if we use it, how many times to use it.
- We can use a recursive approach to try all possible combinations of special offers and regular purchases.
- Memoization is important to avoid redundant calculations, as there can be many overlapping subproblems.
String Problems
Learning Objective
Apply string manipulation concepts to solve a real-world problem.
Shopping Offers
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][0], special[i][1], ..., special[i][n-1] represents the number of pieces of each item and special[i][n] represents the price of the offer.
Return the lowest price you have to pay for exactly certain items as given, where you could make optimal use of the special offers. You are not allowed to buy more items than you want, even if that would lower the overall price. You could use any of the special offers as many times as you want.
Examples
Example 1:
There are two kinds of items, A and B. Their prices are $2 and $5 respectively. In special offer 1, you can pay $5 for 3A and 0B. In special offer 2, you can pay $10 for 1A and 2B. You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.
Example 2:
The price of A is $2, and $3 for B, and $4 for C. You may pay $4 for 1A and 1B, and $9 for 2A, 2B and 1C. You need to buy 1A, 2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, and $4 for 1C. You cannot use special offer #2, since it would make you buy 2A, 2B and 1C, which exceeds your needs.
Example 3:
You don't need to buy anything, so the total price is $0.
Problem Breakdown
Constraints:
- n == price.length
- n == needs.length
- 1 <= n <= 6
- 0 <= price[i] <= 10
- 0 <= needs[i] <= 10
- 1 <= special.length <= 100
- special[i].length == n + 1
- 0 <= special[i][j] <= 50
Key Insights:
This problem can be solved using dynamic programming or recursion with memoization.
For each special offer, we need to decide whether to use it or not, and if we use it, how many times to use it.
We can use a recursive approach to try all possible combinations of special offers and regular purchases.
Memoization is important to avoid redundant calculations, as there can be many overlapping subproblems.
Real-World Application
This problem has several practical applications:
Retail Pricing
Optimizing shopping cart costs by considering various discount offers and bundle deals.
Coupon Management
Finding the best combination of coupons and discounts to minimize the total cost of a purchase.
Budget Planning
Planning purchases to maximize value while staying within budget constraints.