How do you solve Reveal Cards In Increasing Order on LeetCode?

Reveal Cards In Increasing Order (LeetCode #950) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting is essential for revealing cards in increasing order.

What is the time complexity of Reveal Cards In Increasing Order?

The optimal solution for Reveal Cards In Increasing Order runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the result and the queue.

What is the brute force solution for Reveal Cards In Increasing Order?

The brute force approach for Reveal Cards In Increasing Order is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves sorting the deck and simulating the revealing process. Although straightforward, this method is inefficient due to repeated operations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Reveal Cards In Increasing Order?

Reveal Cards In Increasing Order uses the following concepts: Array, Queue, Sorting, Simulation. The recommended patterns to study are: Queue, Sorting.

What are the interview tips for Reveal Cards In Increasing Order?

Always think about how to optimize your approach after the brute force solution. Practice simulating processes with data structures like queues or stacks. Clarify the problem statement and examples with the interviewer to avoid misunderstandings.

What are common mistakes when solving Reveal Cards In Increasing Order?

Not considering the order of operations when revealing cards. Overcomplicating the simulation process instead of using a queue.

#950

Reveal Cards In Increasing Order

Medium

ArrayQueueSortingSimulationQueueSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal approach involves reconstructing the deck in reverse order based on the positions of the sorted cards. This method is efficient and avoids unnecessary operations.

⚙️

Algorithm

3 steps

1Step 1: Sort the deck in ascending order.
2Step 2: Initialize an array of the same length as the deck to hold the result.
3Step 3: Use a queue to simulate the position of cards as they are revealed and moved to the bottom.

solution.py10 lines

1def revealCards(deck):
2    n = len(deck)
3    deck.sort()
4    result = [0] * n
5    queue = deque(range(n))
6    for card in deck:
7        result[queue.popleft()] = card
8        if queue:
9            queue.append(queue.popleft())
10    return result

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the result and the queue.

1Sorting is essential for revealing cards in increasing order.
2Simulating the reveal and move-to-bottom process can be efficiently managed with a queue.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.