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Shopping Offers

Difficulty: Medium

Problem Description

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.

Key Insights

  • Dynamic programming or backtracking can be used to explore all combinations of item purchases.
  • Special offers can be applied multiple times, making it necessary to consider combinations of offers.
  • The problem can be modeled using recursive function with memoization to optimize repeated calculations.

Space and Time Complexity

Time Complexity: O(n * m), where n is the number of items and m is the number of special offers.
Space Complexity: O(n), for storing the state of needs in the recursion stack or memoization.

Solution

The problem can be solved using a recursive function with memoization. The recursive function calculates the minimum cost to fulfill the needs by comparing the cost of purchasing items individually versus using special offers. The base case occurs when there are no remaining needs, returning a cost of zero. For each recursive call, the function checks all available special offers. If applying an offer does not exceed the remaining needs, it recursively calculates the minimum cost for the remaining items after applying the offer. The results are stored in a memoization table to avoid redundant calculations, significantly improving efficiency.

Code Solutions

def shoppingOffers(price, special, needs):
    from functools import lru_cache

    @lru_cache(None)
    def dfs(remaining):
        if all(x == 0 for x in remaining):
            return 0
        
        # Cost if we buy items individually
        cost = sum(p * r for p, r in zip(price, remaining))

        # Check each special offer
        for offer in special:
            new_remaining = tuple(max(0, r - o) for r, o in zip(remaining, offer[:-1]))
            cost = min(cost, offer[-1] + dfs(new_remaining))
        
        return cost

    return dfs(tuple(needs))
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