LeetCodee

950. Reveal Cards In Increasing Order

Jump to Solution: Python Java C++ JavaScript C#

Problem Description

You are given an integer array deck. There is a deck of cards where every card has a unique integer. The integer on the ith card is deck[i].

You can order the deck in any order you want. Initially, all the cards start face down (unrevealed) in one deck.

You will do the following steps repeatedly until all cards are revealed:

  1. Take the top card of the deck, reveal it, and take it out of the deck.
  2. If there are still cards in the deck, put the next top card of the deck at the bottom of the deck.
  3. If there are still unrevealed cards, go back to step 1. Otherwise, stop.

Return an ordering of the deck that would reveal the cards in increasing order.

Example 1:

Input: deck = [17,13,11,2,3,5,7]
Output: [2,13,3,11,5,17,7]
Explanation:
We get the deck in the order [17,13,11,2,3,5,7] (this order does not matter), and reorder it.
After reordering, the deck starts as [2,13,3,11,5,17,7], where 2 is the top of the deck.
We reveal 2, and move 13 to the bottom.  The deck is now [3,11,5,17,7,13].
We reveal 3, and move 11 to the bottom.  The deck is now [5,17,7,13,11].
We reveal 5, and move 17 to the bottom.  The deck is now [7,13,11,17].
We reveal 7, and move 13 to the bottom.  The deck is now [11,17,13].
We reveal 11, and move 17 to the bottom.  The deck is now [13,17].
We reveal 13, and move 17 to the bottom.  The deck is now [17].
We reveal 17.
Since all the cards revealed are in increasing order, the answer is correct.

Example 2:

Input: deck = [1,1000]
Output: [1,1000]

Constraints:

  • 1 <= deck.length <= 1000
  • 1 <= deck[i] <= 10^6
  • All the values of deck are unique.

Solution

Python Solution

class Solution:
    def deckRevealedIncreasing(self, deck: List[int]) -> List[int]:
        n = len(deck)
        deck.sort()
        result = [0] * n
        queue = deque(range(n))
        
        for card in deck:
            result[queue.popleft()] = card
            if queue:
                queue.append(queue.popleft())
        
        return result

Time Complexity: O(n log n)

Where n is the length of the deck. The sorting operation takes O(n log n) time.

Space Complexity: O(n)

We need to store the result array and the queue.

Java Solution

class Solution {
    public int[] deckRevealedIncreasing(int[] deck) {
        int n = deck.length;
        Arrays.sort(deck);
        int[] result = new int[n];
        Queue queue = new LinkedList<>();
        
        for (int i = 0; i < n; i++) {
            queue.offer(i);
        }
        
        for (int card : deck) {
            result[queue.poll()] = card;
            if (!queue.isEmpty()) {
                queue.offer(queue.poll());
            }
        }
        
        return result;
    }
}

Time Complexity: O(n log n)

Where n is the length of the deck. The sorting operation takes O(n log n) time.

Space Complexity: O(n)

We need to store the result array and the queue.

C++ Solution

class Solution {
public:
    vector deckRevealedIncreasing(vector& deck) {
        int n = deck.size();
        sort(deck.begin(), deck.end());
        vector result(n);
        queue q;
        
        for (int i = 0; i < n; i++) {
            q.push(i);
        }
        
        for (int card : deck) {
            result[q.front()] = card;
            q.pop();
            if (!q.empty()) {
                q.push(q.front());
                q.pop();
            }
        }
        
        return result;
    }
};

Time Complexity: O(n log n)

Where n is the length of the deck. The sorting operation takes O(n log n) time.

Space Complexity: O(n)

We need to store the result array and the queue.

JavaScript Solution

/**
 * @param {number[]} deck
 * @return {number[]}
 */
var deckRevealedIncreasing = function(deck) {
    const n = deck.length;
    deck.sort((a, b) => a - b);
    const result = new Array(n).fill(0);
    const queue = [];
    
    for (let i = 0; i < n; i++) {
        queue.push(i);
    }
    
    for (const card of deck) {
        result[queue.shift()] = card;
        if (queue.length > 0) {
            queue.push(queue.shift());
        }
    }
    
    return result;
};

Time Complexity: O(n log n)

Where n is the length of the deck. The sorting operation takes O(n log n) time.

Space Complexity: O(n)

We need to store the result array and the queue.

C# Solution

public class Solution {
    public int[] DeckRevealedIncreasing(int[] deck) {
        int n = deck.Length;
        Array.Sort(deck);
        int[] result = new int[n];
        Queue queue = new Queue();
        
        for (int i = 0; i < n; i++) {
            queue.Enqueue(i);
        }
        
        foreach (int card in deck) {
            result[queue.Dequeue()] = card;
            if (queue.Count > 0) {
                queue.Enqueue(queue.Dequeue());
            }
        }
        
        return result;
    }
}

Time Complexity: O(n log n)

Where n is the length of the deck. The sorting operation takes O(n log n) time.

Space Complexity: O(n)

We need to store the result array and the queue.