A knapsack has a maximum capacity C and there are n items each with weight w[i] and value v[i]. Maximize the knapsack value without exceeding capacity.

The knapsack function solves the 0/1 knapsack problem using a recursive approach with memoization. The function takes as input a list of weights weight, a list of corresponding values values, and a capacity cap. It uses a helper function knapsack_cache that employs a cache to store intermediate results and avoid redundant calculations.

Approach:

1. Base Cases:

   • If there are no items (len(weight) == 0) or if the capacity is zero or negative (cap <= 0), the function returns 0.
   • If there is only one item, the function checks if its weight is less than or equal to the capacity. If yes, it returns the item's value; otherwise, it returns 0.

2. Recursive Case:

   • If the weight of the current item (weight[0]) exceeds the capacity, the function recursively calls itself without including this item.
   • Otherwise, it calculates two values:
     • added_item_0: the value if the current item is included (current item's value plus the result of the recursive call with the remaining items and reduced capacity).
     • remove_item_0: the value if the current item is excluded (result of the recursive call with the remaining items and the same capacity).
   • The function returns the maximum of these two values.

3. Memoization:

   • Before performing any calculations, the function checks if the current state (weight, values, cap) is in the cache. If a match is found, it returns the cached value.
   • After computing a new result, it adds the current state and result to the cache.

Solution

This solution effectively uses memoization to optimize the recursive approach for the 0/1 knapsack problem, ensuring that each state is only computed once and reused when needed, leading to efficient computation in terms of both time and space.

Time Complexity: The time complexity of this approach is $O(n \cdot c)$, where $n$ is the number of items and $c$ is the capacity. This is because there are $n$ items and up to $c$ different capacity values, and each unique state (combination of remaining items and capacity) is computed at most once due to memoization.

Space Complexity: The space complexity is also $O(n \cdot c)$ due to the cache storing intermediate results for each unique state. Additionally, the recursion depth can go up to $O(n)$, but this is dominated by the space required for the cache.