Problem Description

You are given a 0-indexed integer array prices where prices[i] represents the number of coins needed to purchase the (i + 1)-th fruit.

The fruit market has a special reward system:

• When you purchase the (i + 1)-th fruit for prices[i] coins, you can get any number of the next i fruits for free.

For example:

• If you buy the 1st fruit (index 0), you get 0 fruits for free
• If you buy the 2nd fruit (index 1), you can get the next 1 fruit for free
• If you buy the 3rd fruit (index 2), you can get the next 2 fruits for free
• And so on...

An important note: Even if you can take a fruit for free, you still have the option to purchase it with coins to activate its reward and get subsequent fruits for free.

Your task is to find the minimum number of coins needed to acquire all the fruits in the array.

The strategy involves deciding which fruits to buy with coins and which to take for free, optimizing the total cost while ensuring all fruits are obtained.

Intuition

Let's think about this problem step by step. When we buy a fruit at position i, we get the next i fruits for free. This creates overlapping choices - we need to decide which fruits to buy and which to take for free.

The key insight is that we must buy at least some fruits to eventually cover all fruits in the array. For any fruit we buy, we need to consider: what's the optimal way to cover the remaining fruits after taking the free ones?

Consider buying fruit at index i (the (i+1)-th fruit). After buying it:

• We pay prices[i] coins
• We can take the next i+1 fruits for free (since it's the (i+1)-th fruit)
• This covers fruits from index i+1 to index 2i+1

But here's the crucial decision: among the fruits we can take for free, we might want to buy some of them anyway to get their free fruit benefits for covering fruits further ahead.

This naturally leads to a recursive structure: for each fruit position i, we need to find the minimum cost to cover all fruits starting from position i. When we buy fruit i, we then need to decide which fruit to buy next from the range we could take for free.

The base case becomes clear: if buying fruit i gives us enough free fruits to reach the end of the array (when i * 2 >= n), we only need to pay for that fruit and we're done.

Otherwise, we buy fruit i and then recursively find the minimum cost by trying all possible next purchase positions within our free range [i+1, 2i+1]. The formula becomes: cost(i) = prices[i-1] + min(cost(j)) for all valid j in the next purchase range.

Since we'll encounter the same subproblems multiple times (different paths might lead to buying the same fruit position), memoization helps us avoid redundant calculations.

Learn more about Queue, Dynamic Programming, Monotonic Queue and Heap (Priority Queue) patterns.

Solution Approach

The solution uses memoization search (top-down dynamic programming) to find the minimum cost to acquire all fruits.

We implement a recursive function dfs(i) that represents the minimum number of coins needed to buy all fruits starting from the i-th fruit (1-indexed). The answer to our problem is dfs(1).

Implementation Details:

1. Function Definition: dfs(i) computes the minimum cost to cover all fruits from position i onwards.

2. Base Case:

   • When i * 2 >= len(prices), buying the i-th fruit gives us enough free fruits to reach the end
   • We return prices[i - 1] (converting to 0-indexed array access)
   • This happens because buying the i-th fruit gives us i free fruits, so we can cover up to position 2i

3. Recursive Case:

   • We must buy fruit i, which costs prices[i - 1]
   • After buying it, we can take fruits from position i + 1 to 2i for free
   • But we need to buy at least one more fruit to continue covering the rest
   • We try all possible next purchase positions j in the range [i + 1, 2i + 1]
   • The formula becomes: dfs(i) = prices[i - 1] + min(dfs(j) for j in range(i + 1, i * 2 + 2))
   • Note: range(i + 1, i * 2 + 2) generates values from i + 1 to 2i + 1 inclusive

4. Memoization:

   • The @cache decorator automatically memoizes the results
   • This prevents recalculating dfs(i) for the same i value multiple times
   • Without memoization, the time complexity would be exponential

5. Starting Point:

   • We call dfs(1) to find the minimum cost starting from the first fruit
   • Using 1-based indexing in the recursion makes the logic cleaner since the i-th fruit gives i free fruits

The algorithm explores all valid purchasing strategies while avoiding redundant calculations through memoization, ensuring we find the optimal solution efficiently.

Example Walkthrough

Let's walk through a small example with prices = [3, 1, 2] to illustrate the solution approach.

We need to find the minimum cost to acquire all 3 fruits. Let's trace through the recursive calls:

Step 1: Start with dfs(1)

• We're buying the 1st fruit (index 0), which costs 3 coins
• Buying the 1st fruit gives us 1 free fruit (covers position 2)
• We still need to cover position 3
• Next purchase options: positions 2 or 3 (range [2, 3])
• Calculate: dfs(1) = 3 + min(dfs(2), dfs(3))

Step 2: Evaluate dfs(2)

• Buying the 2nd fruit (index 1) costs 1 coin
• Buying the 2nd fruit gives us 2 free fruits (would cover positions 3-4)
• Since 2 * 2 = 4 >= 3 (array length), this reaches the end
• Base case: dfs(2) = 1

Step 3: Evaluate dfs(3)

• Buying the 3rd fruit (index 2) costs 2 coins
• Buying the 3rd fruit gives us 3 free fruits (would cover positions 4-6)
• Since 3 * 2 = 6 >= 3 (array length), this reaches the end
• Base case: dfs(3) = 2

Step 4: Calculate final result

• dfs(1) = 3 + min(1, 2) = 3 + 1 = 4

Verification: The optimal strategy is:

1. Buy fruit 1 for 3 coins (get fruit 2 for free)
2. Buy fruit 2 for 1 coin (get fruits 3 for free)
3. Total cost: 3 + 1 = 4 coins

Alternative strategy (suboptimal):

1. Buy fruit 1 for 3 coins (get fruit 2 for free)
2. Take fruit 2 for free
3. Buy fruit 3 for 2 coins
4. Total cost: 3 + 2 = 5 coins

The algorithm correctly identifies that buying fruit 2 (even though we could take it for free) leads to the minimum total cost of 4 coins.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²)

The recursive function dfs(i) explores different purchasing strategies. For each position i, it considers all possible next positions in the range [i+1, min(2*i+1, n)]. In the worst case, when i is small, this range can contain up to i+1 elements.

• At position 1: we explore up to 2 options
• At position 2: we explore up to 3 options
• At position i: we explore up to i+1 options

Due to memoization with @cache, each state i is computed only once. There are n possible states (positions 1 through n), and for each state, we iterate through at most O(n) next positions in the worst case. Therefore, the overall time complexity is O(n²).

Space Complexity: O(n)

The space complexity consists of:

• The recursion call stack depth: O(n) in the worst case when we traverse from position 1 to position n
• The memoization cache: O(n) for storing results of up to n different positions

Since both components are O(n), the total space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Index Confusion Between 0-based and 1-based Systems

The Pitfall: The most common mistake in this problem is mixing up 0-based array indexing with the 1-based fruit numbering system. The problem states that the (i + 1)-th fruit gives i free fruits, but the array prices uses 0-based indexing. This creates confusion when:

• Accessing the prices array (0-based)
• Calculating how many free fruits you get (1-based logic)
• Determining the range of next possible purchases

Example of the mistake:

# Incorrect implementation mixing indices
def dfs(i):  # If i is 0-based
    if i * 2 >= len(prices):  # Wrong! This uses 0-based logic
        return prices[i]
  
    min_cost = float('inf')
    # Wrong range calculation
    for j in range(i + 1, i * 2 + 1):  
        min_cost = min(min_cost, dfs(j))
    return prices[i] + min_cost

The Solution: Stick to one indexing system within the recursive function. The correct approach uses 1-based indexing for the DFS logic:

• dfs(1) represents buying the 1st fruit (at prices[0])
• Buying the i-th fruit (1-based) gives i free fruits
• Access prices using prices[i - 1] to convert to 0-based
• The range of free fruits extends from position i + 1 to 2i

2. Incorrect Range for Next Purchase Options

The Pitfall: After buying fruit at position i, you get the next i fruits for free (positions i+1 to 2i). The mistake is incorrectly calculating which fruits you can choose to buy next.

Example of the mistake:

# Incorrect: Only considering fruits within the free range
for j in range(i + 1, 2 * i + 1):  # Missing the (2i + 1)-th fruit
    min_cost = min(min_cost, dfs(j))

The Solution: You can buy any fruit from position i + 1 to 2i + 1 (inclusive). Even though fruits from i + 1 to 2i can be taken for free, you might want to buy them to activate their rewards. The (2i + 1)-th fruit is the first fruit that must be purchased if you want it (it's not covered by the free range).

3. Forgetting the Base Case Boundary

The Pitfall: Not correctly identifying when buying a single fruit can cover all remaining fruits.

Example of the mistake:

# Incorrect base case
if current_index >= len(prices):  # Wrong condition
    return 0

The Solution: The base case occurs when current_index * 2 >= len(prices). This means buying the fruit at current_index provides enough free fruits to reach the end of the array. At this point, we only need to pay for the current fruit: prices[current_index - 1].