Shuffle a pack of cards and answer the query

Given a pack of 2^N cards (0 ... 2^N - 1), shuffle it in N steps. At step k (0 < k < N) we divide the deck into 2k equal-sized decks. Each one of those decks is reordered by having all the cards that lie on even positions first, followed by all cards that lie on odd positions (the order is preserved in each one of the two subsequences). Now, we are given a key (index). We have to answer the card on that position (0-based indexing).
Examples: 
 

Input : N = 3 (Size = 2^N), Key = 3
Output : 6
Explanation : 
Pack :      0 1 2 3 4 5 6 7
Shuffle 1 : 0 2 4 6|1 3 5 7
Shuffle 2 : 0 4|2 6|1 5|3 7
Card at index 3 : 6

Method 1: We can simply simulate the whole process and find the exact order of the cards after all the N shuffles are done. 
Time Complexity: O(N * 2^N)
Method 2 : 
Let us try to find the binary representation of Key and the final answer and try to spot some observations based on it.
Let N = 3
Below is the table :
Key ANS 
000 000 
001 100 
010 010 
011 110 
100 001 
101 101 
110 011 
111 111
It is clearly visible that the answer is the reverse of a binary representation of Key. 
 

// C++ program to find the card at given index
// after N shuffles
#include <bits/stdc++.h>
using namespace std;

// function to find card at given index
void shuffle(int N, int key)
{

    // Answer will be reversal of N bits from MSB
    unsigned int NO_OF_BITS = N;
    unsigned int reverse_num = 0, temp;

    // Calculating the reverse binary representation
    for (int i = 0; i < NO_OF_BITS; i++) {
        temp = (key & (1 << i));
        if (temp)
            reverse_num |= (1 << ((NO_OF_BITS - 1) - i));
    }

    // Printing the result
    cout << reverse_num;
}

// driver code
int main()
{
    // No. of Shuffle Steps
    int N = 3;

    // Key position
    unsigned int key = 3;

    shuffle(N, key);
    return 0;
}

// Java program to find the card at given index
// after N shuffles
class GFG {
    
    // function to find card at given index
    static void shuffle(int N, int key)
    {
    
        // Answer will be reversal of N bits from MSB
        int NO_OF_BITS = N;
        int reverse_num = 0, temp;
    
        // Calculating the reverse binary representation
        for (int i = 0; i < NO_OF_BITS; i++) {
            temp = (key & (1 << i));
            if (temp>0)
                reverse_num |= (1 << ((NO_OF_BITS - 1) - i));
        }
    
        // Printing the result
        System.out.print(reverse_num);
    }
    
    //Driver code
    public static void main (String[] args)
    {
        
        // No. of Shuffle Steps
        int N = 3;
    
        // Key position
        int key = 3;
    
        shuffle(N, key);
    }
}

// This code is contributed by Anant Agarwal.

# Python3 program to find the card 
# at given index after N shuffles

# Function to find card at given index
def shuffle(N, key):

    # Answer will be reversal
    # of N bits from MSB
    NO_OF_BITS = N
    reverse_num = 0

    # Calculating the reverse binary representation
    for i in range(NO_OF_BITS):
        temp = (key & (1 << i))
        if (temp):
            reverse_num |= (1 << ((NO_OF_BITS - 1) - i))
    
    # Printing the result
    print(reverse_num)

# Driver code

# No. of Shuffle Steps
N = 3

# Key position
key = 3
shuffle(N, key)

# This code is contributed by Anant Agarwal.

// C# program to find the card at given index
// after N shuffles
using System;

class GFG {
    
    // function to find card at given index
    static void shuffle(int N, int key)
    {
     
        // Answer will be reversal of N bits from MSB
        int NO_OF_BITS = N;
        int reverse_num = 0, temp;
     
        // Calculating the reverse binary representation
        for (int i = 0; i < NO_OF_BITS; i++) {
            temp = (key & (1 << i));
            if (temp > 0)
                reverse_num |= (1 << ((NO_OF_BITS - 1) - i));
        }
     
        // Printing the result
        Console.Write(reverse_num);
    }
    
    //Driver code
    public static void Main()
    {
        
        // No. of Shuffle Steps
        int N = 3;
    
        // Key position
        int key = 3;
    
        shuffle(N, key);
    }
}

// This code is contributed by Anant Agarwal.

<script>

// Javascript program to find the card at given index
// after N shuffles

    // function to find card at given index
    function shuffle(N, key)
    {
    
        // Answer will be reversal of N bits from MSB
        let NO_OF_BITS = N;
        let reverse_num = 0, temp;
    
        // Calculating the reverse binary representation
        for (let i = 0; i < NO_OF_BITS; i++) {
            temp = (key & (1 << i));
            if (temp>0)
                reverse_num |= (1 << ((NO_OF_BITS - 1) - i));
        }
    
        // Printing the result
        document.write(reverse_num);
    }
    

// driver program

        // No. of Shuffle Steps
        let N = 3;
    
        // Key position
        let key = 3;
    
        shuffle(N, key);
        
</script>

Output:  

6

Time Complexity - O(n)

Space Complexity - O(1)