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638. Shopping Offers

medium

Minimize the cost to buy exactly the needed quantity of each item, given a list of special bundle offers and per-item prices. Bag-state recursion plus memoization — the classic 'serialize the state as a key' trick.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

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 j-th item in the i-th offer and special[i][n] (i.e., the last integer in the array) is the price of the i-th 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.

Constraints

  • n == price.length == needs.length
  • 1 <= n <= 6
  • 0 <= price[i], needs[i] <= 10
  • 1 <= special.length <= 100
  • special[i].length == n + 1
  • 0 <= special[i][j] <= 50

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

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Output

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Hints

Progressive — try the first before opening the next.

Hint 1

Recurse on the current needs vector. Try (a) buying everything at regular price, (b) using each special offer that doesn't exceed remaining needs.

Hint 2

Memoize on the needs vector — turn it into a tuple/string key.

Hint 3

If any offer leaves a non-negative needs vector, recurse on the reduced needs + offer price.

Hint 4

Take the min across all branches.

Solution approach

Reveal approach

Recursive shopping(needs) returns the minimum cost. Base option: buy at regular prices = sum(needs[i] * price[i]). For each offer in special, check if applying it keeps all need counts >= 0. If so, recurse on the post-offer needs vector and add the offer price; take the minimum. Memoize on the needs vector — serialize it as a tuple or join the integers with a separator for a hash key. Time is O(N * K * sum(needs)) roughly, where N is the size of the offer list and K is the size of the state space (bounded because each need is small); space is the memo size. Because needs[i] <= 10 and n <= 6, total states are bounded at 11^6 = 1.77M, well within practical limits.

Complexity

Time
O(states * offers * n)
Space
O(states)

Related patterns

  • recursion
  • memoization
  • state-compression

Related problems

Asked at

Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).

  • Google

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