Shopping Offers - Solution & Explanation

This approach uses a combination of Depth First Search (DFS) and memoization to explore all possible ways to purchase the desired items. Memoization is used to store the states that we have already computed, which helps to avoid redundant calculations and speeds up the process.

The idea is to recursively try every combination of buying items using individual purchases and available special offers. For each combination, calculate the remaining needs and the cost so far. If a certain state (combination of needs) has already been computed, retrieve the result from memo to save computation.

The solution defines a recursive function dfs that represents the cost of satisfying the current 'needs'. It uses memoization to save and reuse the results of subproblems, identified by current 'needs' state represented as a tuple.

We calculate the cost of fulfilling the current needs without any special offer. Then, for each offer, we check if that offer is valid by ensuring we don't exceed the current needs. If valid, we recursively calculate the cost using this offer and update the minimum.

This approach employs Dynamic Programming (DP) to solve the problem, considering each combination of remaining needs as a separate subproblem. The DP table (or dictionary/map) stores minimal costs for fulfilling certain needs, calculated iteratively in a bottom-up manner.

Instead of using recursion, this method explicitly builds up solutions for increasingly larger and more complex needs, leveraging already computed simpler results, thereby improving efficiency.

This Java solution utilizes a depth-first approach with dynamic programming to avoid recursively recalculating costs for the same 'needs'. It stores previously calculated results in a hash map that dynamically maintains the cost for a given state of needs.

The cost is initialized with the direct purchase amount of the remaining needs, and iteratively recalculated when checking valid special offers.

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.

First, we convert the shopping list needs into an integer mask, where the quantity of the i-th item needed is stored in the i times 4 to (i + 1) times 4 - 1 bits of mask. For example, when needs = [1, 2, 1], we have mask = 0b0001 0010 0001.

Then, we design a function dfs(cur), representing the minimum amount of money we need to spend when the current state of the shopping list is cur. Therefore, the answer is dfs(mask).

The calculation method of the function dfs(cur) is as follows:

• First, we calculate the cost of the current shopping list cur without using any bundles, denoted as ans.
• Then, we iterate through each bundle offer. If the current shopping list cur can use the bundle offer, i.e., the quantity of each item in cur is not less than that in the bundle offer, then we can try to use this bundle. We subtract the quantity of each item in the bundle offer from cur, obtaining a new shopping list nxt, then recursively calculate the minimum cost of nxt and add the price of the bundle offer[n], updating ans, i.e., ans = min(ans, offer[n] + dfs(nxt)).
• Finally, return ans.

To avoid repeated calculations, we use a hash table f to record the minimum cost corresponding to each state cur.

The time complexity is O(n times k times m^n), where n represents the types of items, and k and m respectively represent the number of bundles and the maximum demand for each type of item. The space complexity is O(n times m^n).