LeetCode 638: Shopping Offers Solution in Python – A Step-by-Step Guide

What Is LeetCode 638: Shopping Offers?

In LeetCode 638: Shopping Offers, you’re given:

• price: List of individual item prices.
• special: List of special offers, each a list of quantities and a total price (e.g., [1, 2, 12] for 1 of item 1, 2 of item 2, $12).
• needs: List of required quantities for each item.

Your task is to compute the minimum cost to buy at least the needed quantities, using any combination of individual purchases and special offers. For example, with price = [2, 5], special = [[3, 0, 5], [1, 2, 10]], and needs = [3, 2], you could buy offers and items to minimize cost. This problem tests your ability to optimize combinations under constraints.

Problem Statement

• Input:
• price: List of integers (item prices).
• special: List of lists (offers: quantities + price).
• needs: List of integers (required quantities).
• Output: An integer—minimum cost to meet or exceed needs.
• Rules:
• Use any number of special offers and individual purchases.
• Must buy at least the quantities in needs.
• Minimize total cost.

Constraints

• 1 ≤ price.length ≤ 6.
• 0 ≤ price[i] ≤ 1000.
• 0 ≤ special.length ≤ 100.
• special[i].length = price.length + 1.
• 0 ≤ special[i][j] ≤ 50 (quantities), 0 ≤ special[i][-1] ≤ 1000 (offer price).
• 0 ≤ needs[i] ≤ 50.

Examples

• Input: price = [2, 5], special = [[3, 0, 5], [1, 2, 10]], needs = [3, 2]
• Output: 14 (e.g., 1 offer [1, 2, 10] + 2 item 1 at $2 each = 14)
• Input: price = [2, 3, 4], special = [[1, 1, 0, 4], [2, 2, 1, 9]], needs = [1, 2, 1]
• Output: 11 (e.g., 1 offer [2, 2, 1, 9] + 1 item 1 at $2 = 11)
• Input: price = [1], special = [], needs = [1]
• Output: 1 (Buy 1 item at $1)

These examples set the cart—let’s solve it!

Understanding the Problem: Minimizing Shopping Cost

To solve LeetCode 638: Shopping Offers in Python, we need to find the cheapest way to buy at least the quantities in needs using individual prices and special offers. A brute-force approach—trying all possible combinations—would be exponential (O(2^(m+n))), where m is offers and n is items, impractical for m ≤ 100. Since offers can be used multiple times, we can optimize with DP or recursion. We’ll use:

• Best Solution (DP with Memoization): O(n m S) time, O(S) space—fast and efficient (n = items, m = offers, S = state space).
• Alternative Solution (Brute-Force Recursion): O(2^(m+n)) time, O(n) space—simple but slow.

Let’s start with the DP solution, breaking it down for beginners.

Best Solution: Dynamic Programming with Memoization

Why This Is the Best Solution

The DP with memoization approach is the top choice for LeetCode 638 because it’s efficient—O(n * m * S) time with O(S) space—and avoids redundant calculations by caching results for each unique needs state. It fits constraints (n ≤ 6, m ≤ 100, needs[i] ≤ 50) and elegantly handles the combinatorial explosion of offer combinations. It’s like a smart shopper memoizing the best deals for every cart state!

How It Works

Think of this as a deal finder:

• Step 1: Define State:
• State = tuple of remaining needs (e.g., (3, 2) for 3 of item 1, 2 of item 2).
• DP(state): Min cost to satisfy state.
• Step 2: Base Case:
• If needs = [0, 0, ...], cost = 0.
• Step 3: Recurrence:
• For each state:
• Option 1: Buy items individually (cost = sum(needs[i] * price[i])).
• Option 2: Apply each offer, recurse on reduced needs.
• Take min of all options.
• Step 4: Memoize:
• Cache results in a dictionary.
• Step 5: Return Result:
• Min cost for initial needs.

It’s like a bargain hunter—try and remember!

Step-by-Step Example

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

• Step 1: State = (3, 2), memo = {}.
• Step 2: DP(3, 2):
• Individual: 3 2 + 2 5 = 16.
• Offer [3, 0, 5]: Needs = [0, 2], cost = 5 + DP(0, 2).
• DP(0, 2): 0 2 + 2 5 = 10.
• Total = 5 + 10 = 15.
• Offer [1, 2, 10]: Needs = [2, 0], cost = 10 + DP(2, 0).
• DP(2, 0): 2 2 + 0 5 = 4.
• Total = 10 + 4 = 14 (best).
• Step 3: Memo[(3, 2)] = 14.
• Output: 14.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def shoppingOffers(self, price: List[int], special: List[List[int]], needs: List[int]) -> int:
        # Step 1: Initialize memoization dictionary
        memo = {}

        # Step 2: Define DP function
        def dp(needs):
            needs_tuple = tuple(needs)  # Convert to tuple for hashing
            if needs_tuple in memo:
                return memo[needs_tuple]

            # Base case: all needs met
            if all(n == 0 for n in needs):
                return 0

            # Option 1: Buy items individually
            cost = sum(n * p for n, p in zip(needs, price))

            # Option 2: Try each special offer
            for offer in special:
                new_needs = needs.copy()
                valid = True
                # Check if offer can be applied
                for i in range(len(needs)):
                    if new_needs[i] < offer[i]:
                        valid = False
                        break
                    new_needs[i] -= offer[i]
                if valid:
                    cost = min(cost, offer[-1] + dp(new_needs))

            memo[needs_tuple] = cost
            return cost

        # Step 3: Start DP with initial needs
        return dp(needs)

• Line 4: Init memo dictionary.
• Lines 7-28: DP function:
• Memoize state, handle base case.
• Compute individual cost, try offers, take min.
• Line 31: Start with initial needs.
• Time Complexity: O(n m S)—n items, m offers, S = state space (bounded by needs).
• Space Complexity: O(S)—memo size.

This is like a deal optimizer—memoize and save!

Alternative Solution: Brute-Force Recursion

Why an Alternative Approach?

The brute-force recursive approach explores all combinations without memoization—O(2^(m+n)) time, O(n) space. It’s simpler, trying every option recursively, but exponentially slow for large inputs (m ≤ 100). It’s like a shopper testing every cart combo!

How It Works

Picture this as a deal tester:

• Step 1: Define recursive function:
• Cost(needs): Min cost for current needs.
• Step 2: Base case: All needs = 0, cost = 0.
• Step 3: Try individual buys and offers, recurse.
• Step 4: Return min cost.

It’s like a recursive bargain hunter—try everything!

Step-by-Step Example

Example: Same as above

• Step 1: Cost(3, 2):
• Individual: 16.
• Offer [3, 0, 5]: Cost(0, 2) = 10, total = 15.
• Offer [1, 2, 10]: Cost(2, 0) = 4, total = 14.
• Step 2: Min(16, 15, 14) = 14.
• Output: 14.

Code for Brute-Force Approach

Copy codeclass Solution:
    def shoppingOffers(self, price: List[int], special: List[List[int]], needs: List[int]) -> int:
        # Step 1: Define recursive function
        def cost(needs):
            if all(n == 0 for n in needs):
                return 0

            # Individual cost
            min_cost = sum(n * p for n, p in zip(needs, price))

            # Try each offer
            for offer in special:
                new_needs = needs.copy()
                valid = True
                for i in range(len(needs)):
                    if new_needs[i] < offer[i]:
                        valid = False
                        break
                    new_needs[i] -= offer[i]
                if valid:
                    min_cost = min(min_cost, offer[-1] + cost(new_needs))

            return min_cost

        # Step 2: Start recursion
        return cost(needs)

• Lines 5-20: Recursive function:
• Base case, individual cost, try offers.
• Time Complexity: O(2^(m+n))—exponential branching.
• Space Complexity: O(n)—recursion stack.

It’s a brute shopper—simple but slow!

Comparing the Two Solutions

• DP with Memoization (Best):
• Pros: O(n m S) time, O(S) space, fast and scalable.
• Cons: Memoization setup.
• Brute-Force (Alternative):
• Pros: O(2^(m+n)) time, O(n) space, straightforward.
• Cons: Exponentially slow.

DP wins for efficiency.

Additional Examples and Edge Cases

• Input: price = [1], special = [], needs = [1]
• Output: 1.
• Input: price = [2, 2], special = [[1, 1, 1]], needs = [2, 2]
• Output: 4 (2 offers).

Both handle these well.

Complexity Breakdown

• DP: Time O(n m S), Space O(S).
• Brute-Force: Time O(2^(m+n)), Space O(n).

DP is optimal.

Key Takeaways

• DP: Memoized deals—smart!
• Brute-Force: Try all—clear!
• Shopping: Optimization is fun.
• Python Tip: Dicts rock—see Python Basics.

Final Thoughts: Shop Those Offers

LeetCode 638: Shopping Offers in Python is a fun optimization challenge. DP with memoization offers speed and elegance, while brute-force provides a clear baseline. Want more? Try LeetCode 322: Coin Change or LeetCode 377: Combination Sum IV. Ready to save? Head to Solve LeetCode 638 on LeetCode and minimize that cost today!