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Shuffle 2n integers as a1-b1-a2-b2-a3-b3-..bn without using extra space | Set 2
  • Last Updated : 03 Feb, 2021

Given an array arr[] consisting of 2* N elements in the form of { a1, a2, …, aN, b1, b2, …, bN }, the task is to shuffle the array to {a1, b1, a2, b2, …, an, b1} without using extra space.

Examples :

Input: arr[] = { 1, 3, 5, 2, 4, 6 }
Output: 1 2 3 4 5 6
Explanation: 
The output contains the elements in the form of { a1, b1, a2, b2, a3, b3 }.

Input: arr[] = {1, 2, 3, -1, -2, -3, }
Output: 1 -1 2 -2 3 -3

Divide and Conquer-based Approach: If the size of an array is a power of 2, then follow the article Shuffle 2n integers in format {a1, b1, a2, b2, a3, b3, ……, an, bn} using divide and conquer technique. 
Time Complexity: O(N * log(N)) 
Auxiliary Space: O(1)

Alternate Approach: The above approach will work for all possible values of N by recursively dividing the array such that the length of both halves is even. Follow the steps below to solve the problem:



  • Define a recursive function, say shuffle(start, end).
    • If array length is divisible by 4, then calculate mid-point of the array, say mid = start + (end – start + 1) / 2.
    • Otherwise, mid = start + (end – start + 1) / 2 – 1.
    • Calculate mid-points of both subarrays, say mid1 = start + (mid – start)/2, and mid2 = mid + (end – mid + 1) / 2.
    • Reverse the subarrays in the ranges [mid1, mid2], [mid1, mid-1], and [mid, mid2 – 1].
    • Make recursive calls for subarrays [start, mid – 1] and [mid, end], i.e. shuffle(start, mid – 1) and shuffle(mid, end) respectively.
  • Finally, print the array.

Illustration:

Consider an array arr[] = {a1, a2, a3, b1, b2, b3}:

  1. Split the array into two halves, both of even length, i.e. a1, a2 : a3, b1, b2, b3.
  2. Reverse the mid of first half to mid of 2nd half, i.e. a1, b1 : a3, a2, b2, b3.
  3. Now, reverse the mid of first half to mid of subarray [0, 5], a1, b1 : a3, a2, b2, b3.
  4. Now reverse mid of subarray [0, 5] to mid of 2nd half, a1, b1 : a2, a3, b2, b3.
  5. Recursively call for arrays {a1, b1}, and {a2, a3, b2, b3}.
  6. Now the array {a2, a3, b2, b3} modifies to {a2, b2, a3, b3} after applying the above operations.
  7. Now the arr[] modifies to {a1, b1, a2, b2, a3, b3}.

Below is the implementation of the above approach:

C

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// C program for the above approach
#include <stdio.h>
 
// Function to reverse the array from the
// position 'start' to position 'end'
void reverse(int arr[], int start, int end)
{
 
    // Stores mid of start and end
    int mid = (end - start + 1) / 2;
 
    // Traverse the array in
    // the range [start, end]
    for (int i = 0; i < mid; i++) {
 
        // Stores arr[start + i]
        int temp = arr[start + i];
 
        // Update arr[start + i]
        arr[start + i] = arr[end - i];
 
        // Update arr[end - i]
        arr[end - i] = temp;
    }
    return;
}
 
// Utility function to shuffle the given array
// in the of form {a1, b1, a2, b2, ....an, bn}
void shuffleArrayUtil(int arr[], int start, int end)
{
    int i;
 
    // Stores the length of the array
    int l = end - start + 1;
 
    // If length of the array is 2
    if (l == 2)
        return;
 
    // Stores mid of the { start, end }
    int mid = start + l / 2;
 
    // Divide array into two
    // halves of even length
    if (l % 4) {
 
        // Update mid
        mid -= 1;
    }
 
    // Calculate the mid-points of
    // both halves of the array
    int mid1 = start + (mid - start) / 2;
    int mid2 = mid + (end + 1 - mid) / 2;
 
    // Reverse the subarray made
    // from mid1 to mid2
    reverse(arr, mid1, mid2 - 1);
 
    // Reverse the subarray made
    // from mid1 to mid
    reverse(arr, mid1, mid - 1);
 
    // Reverse the subarray made
    // from mid to mid2
    reverse(arr, mid, mid2 - 1);
 
    // Recursively calls for both
    // the halves of the array
    shuffleArrayUtil(arr, start, mid - 1);
    shuffleArrayUtil(arr, mid, end);
}
 
// Function to shuffle the given array in
// the form of {a1, b1, a2, b2, ....an, bn}
void shuffleArray(int arr[], int N,
                  int start, int end)
{
 
    // Function Call
    shuffleArrayUtil(arr, start, end);
 
    // Print the modified array
    for (int i = 0; i < N; i++)
        printf("%d ", arr[i]);
}
 
// Driver Code
int main()
{
 
    // Given array
    int arr[] = { 1, 3, 5, 2, 4, 6 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Shuffles the given array to the
    // required permutation
    shuffleArray(arr, N, 0, N - 1);
 
    return 0;
}

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Java

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// Java program for the above approach
 
class GFG{
   
// Function to reverse the array from the
// position 'start' to position 'end'
static void reverse(int arr[], int start, int end)
{
 
    // Stores mid of start and end
    int mid = (end - start + 1) / 2;
 
    // Traverse the array in
    // the range [start, end]
    for (int i = 0; i < mid; i++)
    {
 
        // Stores arr[start + i]
        int temp = arr[start + i];
 
        // Update arr[start + i]
        arr[start + i] = arr[end - i];
 
        // Update arr[end - i]
        arr[end - i] = temp;
    }
    return;
}
 
// Utility function to shuffle the given array
// in the of form {a1, b1, a2, b2, ....an, bn}
static void shuffleArrayUtil(int arr[], int start, int end)
{
    int i;
 
    // Stores the length of the array
    int l = end - start + 1;
 
    // If length of the array is 2
    if (l == 2)
        return;
 
    // Stores mid of the { start, end }
    int mid = start + l / 2;
 
    // Divide array into two
    // halves of even length
    if (l % 4 > 0)
    {
 
        // Update mid
        mid -= 1;
    }
 
    // Calculate the mid-points of
    // both halves of the array
    int mid1 = start + (mid - start) / 2;
    int mid2 = mid + (end + 1 - mid) / 2;
 
    // Reverse the subarray made
    // from mid1 to mid2
    reverse(arr, mid1, mid2 - 1);
 
    // Reverse the subarray made
    // from mid1 to mid
    reverse(arr, mid1, mid - 1);
 
    // Reverse the subarray made
    // from mid to mid2
    reverse(arr, mid, mid2 - 1);
 
    // Recursively calls for both
    // the halves of the array
    shuffleArrayUtil(arr, start, mid - 1);
    shuffleArrayUtil(arr, mid, end);
}
 
// Function to shuffle the given array in
// the form of {a1, b1, a2, b2, ....an, bn}
static void shuffleArray(int arr[], int N,
                  int start, int end)
{
 
    // Function Call
    shuffleArrayUtil(arr, start, end);
 
    // Print the modified array
    for (int i = 0; i < N; i++)
        System.out.printf("%d ", arr[i]);
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given array
    int arr[] = { 1, 3, 5, 2, 4, 6 };
 
    // Size of the array
    int N = arr.length;
 
    // Shuffles the given array to the
    // required permutation
    shuffleArray(arr, N, 0, N - 1);
 
}
}
 
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program for the above approach
 
# Function to reverse the array from the
# position 'start' to position 'end'
def reverse(arr, start, end):
 
    # Stores mid of start and end
    mid = (end - start + 1) // 2
 
    # Traverse the array in
    # the range [start, end]
    for i in range(mid):
 
        # Stores arr[start + i]
        temp = arr[start + i]
 
        # Update arr[start + i]
        arr[start + i] = arr[end - i]
 
        # Update arr[end - i]
        arr[end - i] = temp
    return arr
 
# Utility function to shuffle the given array
# in the of form {a1, b1, a2, b2, ....an, bn}
def shuffleArrayUtil(arr, start, end):
    i = 0
 
    # Stores the length of the array
    l = end - start + 1
 
    # If length of the array is 2
    if (l == 2):
        return
 
    # Stores mid of the { start, end }
    mid = start + l // 2
 
    # Divide array into two
    # halves of even length
    if (l % 4):
 
        # Update mid
        mid -= 1
 
    # Calculate the mid-points of
    # both halves of the array
    mid1 = start + (mid - start) // 2
    mid2 = mid + (end + 1 - mid) // 2
 
    # Reverse the subarray made
    # from mid1 to mid2
    arr = reverse(arr, mid1, mid2 - 1)
 
    # Reverse the subarray made
    # from mid1 to mid
    arr = reverse(arr, mid1, mid - 1)
 
    # Reverse the subarray made
    # from mid to mid2
    arr = reverse(arr, mid, mid2 - 1)
 
    # Recursively calls for both
    # the halves of the array
    shuffleArrayUtil(arr, start, mid - 1)
    shuffleArrayUtil(arr, mid, end)
 
# Function to shuffle the given array in
# the form of {a1, b1, a2, b2, ....an, bn}
def shuffleArray(arr, N, start, end):
 
    # Function Call
    shuffleArrayUtil(arr, start, end)
 
    # Prthe modified array
    for i in arr:
        print(i, end=" ")
 
# Driver Code
if __name__ == '__main__':
 
    # Given array
    arr = [1, 3, 5, 2, 4, 6]
 
    # Size of the array
    N = len(arr)
 
    # Shuffles the given array to the
    # required permutation
    shuffleArray(arr, N, 0, N - 1)
 
# This code is contributed by mohit kumar 29

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C#

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// C# program for the above approach
using System;
public class GFG{
 
// Function to reverse the array from the
// position 'start' to position 'end'
static void reverse(int[] arr, int start, int end)
{
 
    // Stores mid of start and end
    int mid = (end - start + 1) / 2;
 
    // Traverse the array in
    // the range [start, end]
    for (int i = 0; i < mid; i++)
    {
 
        // Stores arr[start + i]
        int temp = arr[start + i];
 
        // Update arr[start + i]
        arr[start + i] = arr[end - i];
 
        // Update arr[end - i]
        arr[end - i] = temp;
    }
    return;
}
 
// Utility function to shuffle the given array
// in the of form {a1, b1, a2, b2, ....an, bn}
static void shuffleArrayUtil(int[] arr, int start, int end)
{
   
    // Stores the length of the array
    int l = end - start + 1;
 
    // If length of the array is 2
    if (l == 2)
        return;
 
    // Stores mid of the { start, end }
    int mid = start + l / 2;
 
    // Divide array into two
    // halves of even length
    if (l % 4 > 0)
    {
 
        // Update mid
        mid -= 1;
    }
 
    // Calculate the mid-points of
    // both halves of the array
    int mid1 = start + (mid - start) / 2;
    int mid2 = mid + (end + 1 - mid) / 2;
 
    // Reverse the subarray made
    // from mid1 to mid2
    reverse(arr, mid1, mid2 - 1);
 
    // Reverse the subarray made
    // from mid1 to mid
    reverse(arr, mid1, mid - 1);
 
    // Reverse the subarray made
    // from mid to mid2
    reverse(arr, mid, mid2 - 1);
 
    // Recursively calls for both
    // the halves of the array
    shuffleArrayUtil(arr, start, mid - 1);
    shuffleArrayUtil(arr, mid, end);
}
 
// Function to shuffle the given array in
// the form of {a1, b1, a2, b2, ....an, bn}
static void shuffleArray(int[] arr, int N,
                  int start, int end)
{
 
    // Function Call
    shuffleArrayUtil(arr, start, end);
 
    // Print the modified array
    for (int i = 0; i < N; i++)
        Console.Write(arr[i] + " ");
}
 
// Driver Code
static public void Main ()
{
 
    // Given array
    int[] arr = { 1, 3, 5, 2, 4, 6 };
 
    // Size of the array
    int N = arr.Length;
 
    // Shuffles the given array to the
    // required permutation
    shuffleArray(arr, N, 0, N - 1);
 
}
}
 
// This code is contributed by Dharanendra L V

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Output: 

1 2 3 4 5 6

 

Time Complexity: O(N * log(N))
Auxiliary Space: O(1)

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