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In Leetcode problem 638, “Shopping Offers,” we have the following scenario:

In a supermarket, there are multiple items which you need to buy, and each item has a different price. Additionally, there are special offers that provide a discounted price for a combination of specific items.

You need to write a program to determine the minimum cost to buy exactly the desired quantities of each item.

The input is given in the following format:

  1. price: A list of integers where price[i] is the price of the i-th item.
  2. special: A list of special offers, where each offer is represented as a list of integers.
    • The first n integers in each offer represent the quantities of items the offer covers.
    • The last integer is the price of the offer.
  3. needs: A list of integers representing the quantity of each item you need to buy.

Example

price = [2, 5]
special = [[3, 0, 5], [1, 2, 10]]
needs = [3, 2]

Output: 10

Clarifying Questions

  1. Can the needs list contain zeroes? (Yes, this means you don’t need that item.)
  2. Are special offers always valid for the given items? (Yes, special offers will correspond to the items in price.)
  3. Can the special offers be used partially or must they satisfy all quantities specified? (Special offers must satisfy all quantities specified in the offer.)

Strategy

  1. Recursive + Memoization: Use depth-first search (DFS) to explore all possible ways to satisfy the needs with minimal cost.
  2. State Representation: Use tuples to represent the current state of needs.
  3. Base Case: When all needs are zero, the cost is zero.
  4. Recursive Case:
    1. Try using each special offer, modify the needs accordingly, and compute the cost.
    2. Also compute the cost of satisfying the needs by buying items individually without any special offers.
  5. Memoization: Cache results for states that have already been computed to avoid redundant calculations.

Code

def shoppingOffers(price, special, needs):
    def dfs(current_needs):
        if tuple(current_needs) in memo:
            return memo[tuple(current_needs)]
        
        # Calculate cost without any special offers, just buying items individually
        min_cost = sum(current_needs[i] * price[i] for i in range(len(price)))
        
        for offer in special:
            temp_needs = current_needs.copy()
            valid_offer = True
            
            for i in range(len(price)):
                temp_needs[i] -= offer[i]
                if temp_needs[i] < 0: # Cannot use this special offer
                    valid_offer = False
                    break
            
            if valid_offer:
                min_cost = min(min_cost, offer[-1] + dfs(temp_needs))

        memo[tuple(current_needs)] = min_cost
        return min_cost

    memo = {}
    return dfs(needs)


# Example usage
price = [2, 5]
special = [[3, 0, 5], [1, 2, 10]]
needs = [3, 2]

print(shoppingOffers(price, special, needs))  # Output: 14

Time Complexity

Thus, the time complexity is approximately O((max_needs+1)^n * n).

This solution should be efficient enough for the typical input sizes expected in such problems.

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