#638

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Shopping Offers

Minimize the cost of purchasing items using available special offers with state transition dynamic programming.

array dynamic programming backtracking bit manipulation memoization

Problem Statement

In a store, you have n items, each with a price given by the price array. You need to buy a certain quantity of each item, as specified in the needs array. Additionally, there are special offers, where each offer consists of a specific number of items and a sale price. The problem is to determine the minimum cost to fulfill the requirements in the needs array using the available special offers.

Each offer is represented by an array of size n + 1, where the first n elements represent the number of items included in the offer, and the last element is the price of the offer. The goal is to decide how to combine these offers to minimize the total cost while satisfying the quantities required in the needs array.

Examples

Example 1

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

Output: 14

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

The price of A is $2, and$ 3 for B, $4 for C. You may pay$ 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 $3 for 1B,$ 4 for 1C. You cannot add more items, though only $9 for 2A ,2B and 1C.

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
The input is generated that at least one of special[i][j] is non-zero for 0 <= j <= n - 1.

Solution Approach

State Transition Dynamic Programming

We define a state as a combination of the remaining needs for each item. Using dynamic programming, we transition between states by applying offers, calculating the cost for each transition. The goal is to minimize the cost for each state and eventually reach the state where no more items are needed.

Memoization for Efficiency

Memoization is key to preventing recalculation of previously visited states. By storing results of subproblems, we avoid redundant computations, thus speeding up the solution process, especially when the number of special offers is large.

Backtracking to Explore Combinations

Backtracking is employed to explore different combinations of offers. By recursively trying offers, we can determine the best sequence of selections that minimize the total cost. This method ensures that all combinations are considered without missing any optimal solution.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is influenced by the number of possible states, which grows exponentially based on the needs array, with additional complexity introduced by the number of special offers. Memoization reduces the redundant calculations, improving the efficiency of the approach. The space complexity is determined by the number of states stored in the memoization table, which again depends on the size of the needs array and the number of special offers.

What Interviewers Usually Probe

Check if the candidate can efficiently manage state transitions and handle large numbers of special offers.
Observe how well the candidate applies memoization to avoid redundant calculations.
Assess whether the candidate effectively handles the backtracking approach to explore possible combinations.

Common Pitfalls or Variants

Common pitfalls

Forgetting to account for all possible states during the transition, which may lead to missed optimal solutions.
Inefficient implementation of memoization, leading to excessive recomputation and slow performance.
Failing to consider all combinations of offers, which can lead to suboptimal solutions.

Follow-up variants

How does the approach scale when the number of items increases beyond 6?
What if there were constraints limiting the number of times a single offer can be used?
Can this problem be solved using a greedy approach instead of dynamic programming?

FAQ

What is the main pattern used in Shopping Offers?

The problem uses a state transition dynamic programming pattern, which involves exploring different combinations of special offers while minimizing the total cost.

How does memoization improve the performance of the solution?

Memoization stores the results of previously calculated states, preventing redundant calculations and significantly speeding up the process.

What happens if I don't use backtracking in this problem?

Without backtracking, you may miss some combinations of special offers that could lead to the optimal solution, potentially resulting in higher costs.

Can the solution be optimized further for larger inputs?

Yes, by improving the way states are stored and transitioned, you can optimize the solution for larger inputs, such as reducing space complexity.

What is the space complexity of the solution?

The space complexity depends on the number of possible states in the memoization table, which is influenced by the size of the needs array and the number of special offers.

terminal

Solution

Solution 1: State Compression + Memoization Search

We notice that the number of types of items $n \leq 6$ in the problem, and the quantity of each item needed does not exceed $10$. We can use $4$ binary bits to represent the quantity of each item needed. Thus, we only need at most $6 \times 4 = 24$ binary bits to represent the entire shopping list.

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class Solution:
    def shoppingOffers(
        self, price: List[int], special: List[List[int]], needs: List[int]
    ) -> int:
        @cache
        def dfs(cur: int) -> int:
            ans = sum(p * (cur >> (i * bits) & 0xF) for i, p in enumerate(price))
            for offer in special:
                nxt = cur
                for j in range(len(needs)):
                    if (cur >> (j * bits) & 0xF) < offer[j]:
                        break
                    nxt -= offer[j] << (j * bits)
                else:
                    ans = min(ans, offer[-1] + dfs(nxt))
            return ans

        bits, mask = 4, 0
        for i, need in enumerate(needs):
            mask |= need << i * bits
        return dfs(mask)

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