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Solution Approach

Solution Overview

There are 2 main approaches to solve this problem:

  1. Recursion with Memoization - Complex approach
  2. Dynamic Programming - Complex approach

Approach 1: Recursion with Memoization

The key insight for solving this problem is to use recursion with memoization to explore all possible combinations of special offers and regular purchases.

For each special offer, we need to decide whether to use it or not. If we use it, we need to update our remaining needs and recursively solve the problem for the updated needs.

Let's define a recursive function shoppingOffers(price, special, needs) that returns the minimum price to fulfill the needs.

For each special offer, we check if it's valid (i.e., it doesn't exceed our needs for any item). If it's valid, we update our needs by subtracting the items in the offer, and recursively call the function with the updated needs.

We also consider the option of not using any special offer and buying all items at their regular prices.

The minimum of all these options is our answer.

To avoid redundant calculations, we use memoization to store the results of subproblems.

Algorithm:

  1. Define a recursive function shoppingOffers(price, special, needs) that returns the minimum price to fulfill the needs.
  2. Use a hash map to store the results of subproblems (memoization).
  3. For each special offer, check if it's valid (i.e., it doesn't exceed our needs for any item).
  4. If the offer is valid, update the needs by subtracting the items in the offer, and recursively call the function with the updated needs.
  5. Also consider the option of not using any special offer and buying all items at their regular prices.
  6. Return the minimum of all these options.

Time Complexity:

O(m * n * k)

Where m is the number of special offers, n is the number of items, and k is the maximum number of times a special offer can be used. In the worst case, we need to try all possible combinations of special offers.

Space Complexity:

O(n * k^n)

We use a hash map to store the results of subproblems, and there are O(k^n) possible states (each item can have k different quantities).

Approach 2: Dynamic Programming

We can also solve this problem using dynamic programming. The key is to define a state that represents the current needs and compute the minimum cost to fulfill those needs.

Let's define dp[needs] as the minimum price to fulfill the needs.

For each state, we consider two options:

  1. Buy all items at their regular prices.
  2. Use a special offer and recursively solve the problem for the remaining needs.

The minimum of these two options is our answer for the current state.

Since the state space is large (there are many possible combinations of needs), we use a hash map to store the DP values.

Algorithm:

  1. Define a DP state dp[needs] as the minimum price to fulfill the needs.
  2. Use a hash map to store the DP values.
  3. For each state, consider two options:
  4. a. Buy all items at their regular prices.
  5. b. Use a special offer and recursively solve the problem for the remaining needs.
  6. Return dp[needs] as the minimum price to fulfill the needs.

Time Complexity:

O(m * n * k)

Where m is the number of special offers, n is the number of items, and k is the maximum number of times a special offer can be used. We need to compute the DP value for each possible state.

Space Complexity:

O(n * k^n)

We use a hash map to store the DP values, and there are O(k^n) possible states.

Pseudocode

solution.pseudo
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function shoppingOffers(price, special, needs):
    memo = new HashMap<String, Integer>()
 
    function dfs(needs):
        // Convert needs to a string for memoization
        key = needs.toString()
 
        // If we've already computed the result for this state, return it
        if key in memo:
            return memo[key]
 
        // Option 1: Buy all items at their regular prices
        result = 0
        for i from 0 to needs.length - 1:
            result += needs[i] * price[i]
 
        // Option 2: Use a special offer
        for offer in special:
            // Check if the offer is valid
            valid = true
            for i from 0 to needs.length - 1:
                if needs[i] < offer[i]:
                    valid = false
                    break
 
            if valid:
                // Update the needs
                newNeeds = needs.clone()
                for i from 0 to needs.length - 1:
                    newNeeds[i] -= offer[i]
 
                // Recursively solve the problem for the updated needs
                result = min(result, offer[offer.length - 1] + dfs(newNeeds))
 
        // Memoize and return the result
        memo[key] = result
        return result
 
    return dfs(needs)
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