Problem Description

You are at a fruit market with different types of exotic fruits on display. The fruits are arranged in a line and numbered starting from 1 (1-indexed).

Each fruit has a price, given in an array prices, where prices[i] represents the number of coins needed to purchase the i-th fruit.

The market has a special offer:

• When you purchase the i-th fruit for prices[i] coins, you get the next i fruits for free

For example:

• If you buy fruit 1, you get fruit 2 for free (1 fruit free)
• If you buy fruit 2, you get fruits 3 and 4 for free (2 fruits free)
• If you buy fruit 3, you get fruits 4, 5, and 6 for free (3 fruits free)

Important note: Even if you can take a fruit for free due to a previous purchase, you can still choose to buy it with coins to activate its own offer.

Your goal is to find the minimum number of coins needed to acquire all the fruits in the market.

The problem essentially asks you to strategically decide which fruits to buy with coins (to activate their offers) and which fruits to take for free (from previous offers), such that you end up with all fruits while spending the minimum amount of coins.

Intuition

Let's think about this problem backwards. If we're standing at fruit i, what's the minimum cost to get all fruits from position i to the end?

When we buy fruit i, we get the next i fruits for free. This means if we buy fruit i, we can skip paying for fruits i+1, i+2, ..., i+i. But we still need to ensure all fruits after position i+i are obtained somehow.

This leads us to a key insight: if we buy fruit i, we need to make sure all fruits from position i+i+1 onwards are covered. The way to do this is to buy some fruit j where i < j ≤ 2i+1 (within our free range or just after it) that will help cover the remaining fruits.

So for each fruit i, the minimum cost is: f[i] = prices[i-1] + min(f[j]) where j ranges from i+1 to 2i+1

Working backwards from the last fruit makes sense because:

• The last fruit only costs its price (no fruits after it to worry about)
• Each previous fruit needs to consider the optimal ways to cover all fruits after it

The challenge is that for each position i, we need to find the minimum value among f[i+1], f[i+2], ..., f[2i+1]. As we move from right to left (decreasing i), this range keeps changing - the left boundary increases while the right boundary decreases.

This is essentially a sliding window problem where we need to track the minimum value in a window that's constantly shifting and shrinking. A monotonic deque is perfect for this - it maintains elements in increasing order, allowing us to efficiently:

• Remove elements that fall outside our current window
• Find the minimum value (always at the front)
• Add new elements while maintaining the monotonic property

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

Solution Approach

We implement a dynamic programming solution optimized with a monotonic deque. The solution works backwards from the last fruit to the first.

Core Algorithm:

1. Initialize a deque q to maintain indices of fruits in a way that helps us find the minimum cost efficiently.

2. Iterate backwards from fruit n to fruit 1:

   • For each fruit i, we need to find the minimum cost among fruits in range [i+1, 2i+1]

3. Maintain the sliding window using the deque:

   • Remove outdated elements: While the front of the deque contains indices greater than 2i+1, remove them (they're outside our valid range)
   • Update the current fruit's cost: If i ≤ (n-1)/2, add the minimum cost from the valid range to prices[i-1]. The minimum is always at prices[q[0]-1] due to our monotonic property
   • Maintain monotonic property: Before adding index i to the deque, remove all indices from the back whose corresponding prices are greater than or equal to prices[i-1]. This ensures the deque remains monotonically increasing by price values
   • Add current index: Append i to the deque

Key Implementation Details:

• We reuse the prices array for dynamic programming values, where prices[i-1] ultimately stores the minimum cost to get all fruits starting from fruit i

• The condition i ≤ (n-1)/2 is crucial:

  • When i > (n-1)/2, buying fruit i gives us enough free fruits to cover all remaining fruits (since we get i free fruits and there are at most n-i fruits after position i)
  • When i ≤ (n-1)/2, we need to buy additional fruits later, so we add the minimum cost from our valid range

• The monotonic deque optimization reduces time complexity from O(n²) to O(n) because:

  • Each element is added to the deque once and removed at most once
  • Finding the minimum is O(1) (always at the front of the deque)

The final answer is prices[0], which represents the minimum cost to acquire all fruits starting from fruit 1.

Example Walkthrough

Let's walk through a concrete example with prices = [3, 1, 2] (3 fruits total).

Initial Setup:

• We have 3 fruits with prices: fruit 1 costs 3, fruit 2 costs 1, fruit 3 costs 2
• We'll work backwards from fruit 3 to fruit 1
• Initialize an empty deque q = []

Step 1: Process fruit 3 (i = 3)

• Current window range: [4, 7] (i+1 to 2i+1)
• Since there are no fruits beyond position 3, no additional cost needed
• Check condition: i = 3 > (n-1)/2 = 1, so prices[2] remains 2
• Update deque: Add index 3 to deque → q = [3]
• Current DP state: prices = [3, 1, 2]

Step 2: Process fruit 2 (i = 2)

• Current window range: [3, 5] (i+1 to 2i+1)
• Remove outdated: Front element 3 ≤ 5, so keep it
• Check condition: i = 2 > (n-1)/2 = 1, so prices[1] remains 1
  • This means buying fruit 2 gives us 2 free fruits, covering fruits 3 and 4 (enough to get all remaining fruits)
• Maintain monotonic property: prices[2] = 2 > prices[1] = 1, so remove 3 from back
• Update deque: Add index 2 → q = [2]
• Current DP state: prices = [3, 1, 2]

Step 3: Process fruit 1 (i = 1)

• Current window range: [2, 3] (i+1 to 2i+1)
• Remove outdated: Front element 2 ≤ 3, so keep it
• Check condition: i = 1 ≤ (n-1)/2 = 1, so we need to add minimum cost
  • Minimum in range is at q[0] = 2, so prices[1-1] = 1
  • Update: prices[0] = 3 + 1 = 4
• Maintain monotonic property: prices[1] = 1 < prices[0] = 4, so nothing to remove
• Update deque: Add index 1 → q = [2, 1]
• Final DP state: prices = [4, 1, 2]

Result: The minimum cost is prices[0] = 4

Verification: Let's verify this is correct by checking different strategies:

1. Buy fruit 1 only (cost 3): Get fruit 2 free, but need to buy fruit 3 → Total: 3 + 2 = 5
2. Buy fruits 1 and 2 (cost 3 + 1): Fruit 1 gives fruit 2 free (redundant), fruit 2 gives fruits 3 and 4 free → Total: 4 ✓
3. Buy fruit 2 only (cost 1): Get fruits 3 and 4 free, but fruit 1 isn't covered → Invalid
4. Buy all fruits: 3 + 1 + 2 = 6

The optimal strategy is indeed to buy fruits 1 and 2 for a total cost of 4 coins.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array prices.

The algorithm iterates through the array once in reverse order (from n to 1), performing O(1) amortized operations with the deque in each iteration. While there are nested while loops inside the main loop, each element can be added to and removed from the deque at most once throughout the entire execution. The first while loop removes elements that are out of range, and the second while loop maintains a monotonic property by removing elements. Since each of the n elements is pushed exactly once and popped at most once, the total number of deque operations across all iterations is O(n).

The space complexity is O(n), where n is the length of the array prices.

The primary space usage comes from the deque q, which in the worst case can store up to n elements when all elements satisfy the monotonic property maintained by the algorithm. The algorithm modifies the input array prices in-place, but since this is already provided as input, it doesn't count as additional space. No other data structures scale with the input size.

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

Common Pitfalls

1. Index Confusion Between 0-indexed Array and 1-indexed Problem

The problem describes fruits numbered from 1 (1-indexed), but the prices array is 0-indexed. This creates multiple opportunities for off-by-one errors.

Pitfall Example:

# Incorrect: Mixing up indices
for current_index in range(n, 0, -1):
    while dq and dq[0] > current_index * 2 + 1:  # Using 1-indexed logic
        dq.popleft()
    prices[current_index] += prices[dq[0]]  # Wrong! Array access should be index-1

Solution:

• Consistently track whether you're working with 1-indexed positions (fruit numbers) or 0-indexed positions (array indices)
• When storing indices in the deque, decide on one convention and stick to it
• Add clear comments indicating which indexing system each variable uses

2. Incorrect Sliding Window Range Calculation

When buying fruit i (1-indexed), you get the next i fruits free, meaning fruits [i+1, i+i] are free. The range for finding minimum cost should be [i+1, min(2*i+1, n)].

Pitfall Example:

# Incorrect: Wrong upper bound
while dq and dq[0] > current_index * 2:  # Should be 2*i + 1
    dq.popleft()

# Or forgetting to cap at n
while dq and dq[0] > current_index * 2 + 1:  # Doesn't handle when 2*i+1 > n
    dq.popleft()

Solution:

• The upper bound should be 2*i + 1 (for 1-indexed) or min(2*i + 1, n) if being explicit
• Since we process backwards and maintain the deque properly, indices beyond n naturally won't exist in the deque

3. Forgetting the Special Case for Large Indices

When i > (n-1)/2, buying fruit i gives enough free fruits to cover all remaining fruits, so we don't need to add any additional cost.

Pitfall Example:

# Incorrect: Always adding minimum cost
for current_index in range(n, 0, -1):
    if dq:  # Wrong condition!
        prices[current_index - 1] += prices[dq[0] - 1]

Solution:

• Only add the minimum cost when current_index <= (n-1)//2
• This ensures we don't unnecessarily add costs when the current purchase already covers all remaining fruits

4. Improper Deque Maintenance for Monotonic Property

The deque should maintain indices in increasing order of their corresponding prices to ensure O(1) minimum finding.

Pitfall Example:

# Incorrect: Using wrong comparison
while dq and prices[dq[-1] - 1] > prices[current_index - 1]:  # Should be >=
    dq.pop()
  
# Or comparing indices instead of values
while dq and dq[-1] >= current_index:  # Wrong! Should compare prices
    dq.pop()

Solution:

• Use >= instead of > to handle equal values properly and maintain strict monotonicity
• Always compare the actual price values, not the indices themselves
• This ensures the deque front always contains the index with minimum price in the valid range