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++
// 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 indexvoid 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 codeint main(){ // No. of Shuffle Steps int N = 3; // Key position unsigned int key = 3; shuffle(N, key); return 0;} |
Java
// Java program to find the card at given index// after N shufflesclass 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
# Python3 program to find the card # at given index after N shuffles# Function to find card at given indexdef 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 StepsN = 3# Key positionkey = 3shuffle(N, key)# This code is contributed by Anant Agarwal. |
C#
// C# program to find the card at given index// after N shufflesusing 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. |
Javascript
<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)
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