Problem Description

You have a deck of cards represented as an integer array deck, where each card has a unique integer value. The value of the i-th card is deck[i].

You need to arrange the deck in a specific order so that when you perform a particular revealing process, the cards are revealed in increasing order.

The revealing process works as follows:

1. Take the top card from the deck, reveal it, and remove it from the deck
2. If cards remain in the deck, move the next top card to the bottom of the deck
3. Repeat steps 1-2 until all cards are revealed

Your task is to find the initial arrangement of the deck that will result in the cards being revealed in increasing order (smallest to largest) when following this process.

For example, if you have cards [17, 13, 11, 2, 3, 5, 7], you need to arrange them in such a way that when you:

• Reveal the top card (should be 2)
• Move the next card to bottom
• Reveal the top card (should be 3)
• Move the next card to bottom
• And continue this pattern...

The cards are revealed as: 2, 3, 5, 7, 11, 13, 17 (in increasing order).

The function should return an array representing the initial deck arrangement, where the first element is the top of the deck.

Intuition

The key insight is to work backwards from the desired outcome. We know we want to reveal cards in increasing order: smallest, second smallest, third smallest, and so on. Instead of trying to figure out the initial arrangement directly, we can simulate the revealing process in reverse.

Think about what happens during the forward process:

• We reveal a card (remove from top)
• We move the next top card to the bottom
• Repeat

To reverse this process:

• Start with the largest card (since it's revealed last)
• To "un-move" a card from bottom to top, we take the bottom card and put it on top
• To "un-reveal" a card, we place it on top of the deck

Working backwards with sorted values in descending order:

1. Start with an empty deck
2. For each card value (from largest to smallest):
   • If the deck isn't empty, "undo" the move operation by taking the bottom card and putting it on top
   • Place the current card on top of the deck

This reversal works because each step undoes what would happen in the forward process. When we place the largest card, it will be the last to be revealed. When we place the second-largest card and move a card from bottom to top, we're setting up the deck so that after revealing and moving cards in the forward process, the second-largest will be revealed at the right time.

By building the deck this way from largest to smallest, we ensure that when the forward revealing process is applied, cards will be revealed from smallest to largest.

Learn more about Queue and Sorting patterns.

Solution Approach

The solution implements the reverse simulation approach using a deque (double-ended queue) data structure, which allows efficient operations at both ends.

Implementation Steps:

1. Sort the deck in descending order: First, we sort the input array in reverse order using sorted(deck, reverse=True). This gives us the cards from largest to smallest.

2. Initialize a deque: Create an empty deque q that will hold our final arrangement.

3. Build the deck backwards: For each card value v in the sorted array (from largest to smallest):

   • Undo the "move to bottom" operation: If the deque is not empty, we simulate undoing the move operation by taking the last card (q.pop()) and putting it at the front (q.appendleft()). This reverses the action of moving a card from top to bottom.
   • Place the current card on top: Add the current card v to the front of the deque using q.appendleft(v).

4. Return the result: Convert the deque to a list and return it.

Why deque? The deque data structure is chosen because it provides O(1) time complexity for both:

• appendleft(): Adding elements to the front
• pop(): Removing elements from the end

Example walkthrough: For deck = [17, 13, 11, 2, 3, 5, 7]:

• Sorted descending: [17, 13, 11, 7, 5, 3, 2]
• Process:
  • Add 17: q = [17]
  • Move last to front, add 13: q = [13, 17]
  • Move last to front, add 11: q = [11, 17, 13]
  • Continue this pattern...
• Final result: [2, 13, 3, 11, 5, 17, 7]

This arrangement, when processed with the revealing algorithm, will reveal cards in order: 2, 3, 5, 7, 11, 13, 17.

Example Walkthrough

Let's work through a small example with deck = [1, 2, 3, 4] to understand how the reverse simulation builds the correct initial arrangement.

Goal: We want cards to be revealed in order: 1, 2, 3, 4

Step 1: Sort in descending order

• Sorted deck: [4, 3, 2, 1]
• We'll process these from largest to smallest

Step 2: Build the deck backwards using a deque

Starting with empty deque q = []

Processing 4 (largest):

• Deque is empty, so no move operation to undo
• Add 4 to front: q = [4]

Processing 3:

• Deque is not empty, so undo the move operation:
  • Take last card (4) and put it at front: q = [4] (no change since only one card)
• Add 3 to front: q = [3, 4]

Processing 2:

• Deque is not empty, so undo the move operation:
  • Take last card (4) and put it at front: q = [4, 3]
• Add 2 to front: q = [2, 4, 3]

Processing 1:

• Deque is not empty, so undo the move operation:
  • Take last card (3) and put it at front: q = [3, 2, 4]
• Add 1 to front: q = [1, 3, 2, 4]

Step 3: Verify the result

Let's verify [1, 3, 2, 4] reveals cards in order 1, 2, 3, 4:

Initial deck: [1, 3, 2, 4] (top is leftmost)

1. Reveal top card: 1 | Remaining: [3, 2, 4]
2. Move top to bottom: [2, 4, 3]
3. Reveal top card: 2 | Remaining: [4, 3]
4. Move top to bottom: [3, 4]
5. Reveal top card: 3 | Remaining: [4]
6. No cards to move
7. Reveal top card: 4 | Remaining: []

Cards revealed in order: 1, 2, 3, 4 ✓

The reverse simulation correctly builds the initial arrangement by working backwards from the desired reveal order.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n log n) where n is the length of the deck array.

• Sorting the deck takes O(n log n) time
• The for loop iterates through all n elements once
• Inside the loop, q.pop() and q.appendleft() operations on a deque are O(1) operations
• Therefore, the loop contributes O(n) time
• Overall time complexity: O(n log n) + O(n) = O(n log n)

Space Complexity: O(n) where n is the length of the deck array.

• The sorted operation creates a new sorted list which takes O(n) space
• The deque q stores all n elements from the deck, requiring O(n) space
• The final list conversion also creates a new list of size n, but this is the return value
• Overall space complexity: O(n)

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

Common Pitfalls

1. Attempting Forward Simulation Instead of Reverse

One of the most common mistakes is trying to simulate the revealing process forward rather than backward. Many people initially try to place cards directly in their final positions by simulating the reveal process, which becomes extremely complex and error-prone.

Incorrect Approach:

# Trying to simulate forward - This doesn't work!
def incorrectApproach(deck):
    sorted_deck = sorted(deck)
    result = [0] * len(deck)
    positions = list(range(len(deck)))
  
    for i, card in enumerate(sorted_deck):
        # Try to find where to place this card
        # This becomes very complicated...

Why it fails: The forward simulation requires tracking multiple indices and simulating the exact reveal process while placing cards, which leads to complex index management and off-by-one errors.

2. Forgetting to Handle the Special Case of Single Card

While the provided solution handles this correctly due to the if result_queue: check, a common mistake is not considering edge cases properly.

Potential Bug:

def buggyVersion(deck):
    result_queue = deque()
  
    for card_value in sorted(deck, reverse=True):
        # Missing the check for empty queue!
        result_queue.appendleft(result_queue.pop())  # Error when queue is empty
        result_queue.appendleft(card_value)
  
    return list(result_queue)

3. Using Wrong Data Structure (List Instead of Deque)

Using a regular list instead of deque leads to inefficient O(n) operations for inserting at the beginning.

Inefficient Implementation:

def inefficientVersion(deck):
    result = []
  
    for card_value in sorted(deck, reverse=True):
        if result:
            # O(n) operation - moving last element to front
            result = [result[-1]] + result[:-1]
        # O(n) operation - inserting at beginning
        result = [card_value] + result
  
    return result

Time Complexity Issue: This changes the overall complexity from O(n log n) to O(n²).

4. Incorrect Order of Operations When Building Backwards

A subtle but critical mistake is performing the operations in the wrong order during the reverse simulation.

Wrong Order:

def wrongOrder(deck):
    result_queue = deque()
  
    for card_value in sorted(deck, reverse=True):
        # Adding card first, then moving - This is wrong!
        result_queue.appendleft(card_value)
        if len(result_queue) > 1:
            result_queue.appendleft(result_queue.pop())
  
    return list(result_queue)

Correct Order: First undo the "move to bottom" operation (if needed), then place the current card on top.

Solution to Avoid These Pitfalls:

1. Think in Reverse: Always approach this problem by thinking about undoing the operations rather than simulating forward.

2. Use Proper Data Structures: Stick with deque for O(1) operations at both ends.

3. Trace Through Small Examples: Before coding, manually trace through a small example (like [1, 2, 3]) in reverse to understand the pattern.

4. Validate with the Original Process: After getting your result, verify it by simulating the actual reveal process to ensure cards appear in increasing order.

Verification Helper:

def verify_solution(arranged_deck):
    """Verify that the arranged deck reveals cards in increasing order."""
    queue = deque(arranged_deck)
    revealed = []
  
    while queue:
        # Reveal top card
        revealed.append(queue.popleft())
        # Move next top card to bottom (if exists)
        if queue:
            queue.append(queue.popleft())
  
    # Check if revealed cards are in increasing order
    return revealed == sorted(revealed)