Find Minimum Number of Coins for Fruits II

Given a 1-indexed array prices[], where prices[i] represents the cost of the i_th fruit in a store. The store has a special offer: if you buy the i_th fruit for prices[i] coins, you can get the next i fruit for free.

Note: Even if you can get an fruit j for free, you still have the option to purchase it for prices[j] coins to receive a new offer. The task is to return the minimum number of coins needed to acquire all the fruits.

Examples:

Input: prices = [30, 10, 20]
Output: 40
Explanation:

• Purchase the 1st fruit with 30 coins. Now, you are allowed to take the 2nd fruit for free, total coins = 30
• Purchase the 2nd fruit with 1 coin. Now, you are allowed to take the 3rd fruit for free, total coins = 30 + 10 = 40.
• Take the 3rd fruit for free, total coins = 40.

Input: prices = [2, 20, 2, 2]
Output: 4
Explanation:

• Purchase the 1st fruit with 1 coin. Now, you are allowed to take the 2nd fruit for free, total cost = 2.
• Take the 2nd fruit for free, total cost = 2.
• Purchase the 3rd fruit for 1 coin. Now, you are allowed to take the 4th fruit for free, total cost = 2 + 2.
• Take the 4th fruit for free, total cost = 4.

Approach: To solve the problem, follow the below idea:

The idea is to use the "take" and "nottake" approach of DP with memoization. When we take the i_th fruit , we directly consider moving to fruits from i+1 to 2*i+2, skipping the next i fruits as per the offer. We recursively explore each valid move from these indices and find the minimum cost. Add the price of the current fruit to this minimum cost and return it. To optimize, we use a memoization vector initialized to zero to store the results of subproblems, ensuring we do not recompute values, improving efficiency.

Step-by-step algorithm:

• Initialize a memoization array with default values set to - 1.
• Base Case: If the current index exceeds the length, return 0.
• Recursive Case: For each index, compute the total cost by taking the current fruit's price and recursively solving for the next valid index (i + i + 1).
• If the result is already computed, return it
• Add the current fruit's price to the minimum cost across all possible indices and return it.
• Update the memoization array with the minimum cost for each index as you compute them.

Below is the implementation of the above algorithm:

40
4

Time Complexity: O(n), where n is the number of fruits in the fruit market.
Auxiliary Space: O(n)