Given an array of **2n** elements in the following format { a1, a2, a3, a4, ….., an, b1, b2, b3, b4, …., bn }. The task is shuffle the array to {a1, b1, a2, b2, a3, b3, ……, an, bn } without using extra space.

Examples:

Input : arr[] = { 1, 2, 9, 15 } Output : 1 9 2 15 Input : arr[] = { 1, 2, 3, 4, 5, 6 } Output : 1 4 2 5 3 6

**Method 1: Brute Force**

A brute force solution involves two nested loops to rotate the elements in the second half of the array to the left. The first loop runs n times to cover all elements in the second half of the array. The second loop rotates the elements to the left. Note that the start index in the second loop depends on which element we are rotating and the end index depends on how many positions we need to move to the left.

Below is implementation of this approach:

## C++

// C++ Naive program to shuffle an array of size 2n #include <bits/stdc++.h> using namespace std; // function to shuffle an array of size 2n void shuffleArray(int a[], int n) { // Rotate the element to the left for (int i = 0, q = 1, k = n; i < n; i++, k++, q++) for (int j = k; j > i + q; j--) swap(a[j-1], a[j]); } // Driven Program int main() { int a[] = { 1, 3, 5, 7, 2, 4, 6, 8 }; int n = sizeof(a) / sizeof(a[0]); shuffleArray(a, n/2); for (int i = 0; i < n; i++) cout << a[i] << " "; return 0; }

## Java

// Java Naive program to shuffle an array of size 2n import java.util.Arrays; public class GFG { // method to shuffle an array of size 2n static void shuffleArray(int a[], int n) { // Rotate the element to the left for (int i = 0, q = 1, k = n; i < n; i++, k++, q++) for (int j = k; j > i + q; j--){ // swap a[j-1], a[j] int temp = a[j-1]; a[j-1] = a[j]; a[j] = temp; } } // Driver Method public static void main(String[] args) { int a[] = { 1, 3, 5, 7, 2, 4, 6, 8 }; shuffleArray(a, a.length/2); System.out.println(Arrays.toString(a)); } }

## Python3

# Python3 Naive program to # shuffle an array of size 2n # Function to shuffle an array of size 2n def shuffleArray(a, n): # Rotate the element to the left i, q, k = 0, 1, n while(i < n): j = k while(j > i + q): a[j - 1], a[j] = a[j], a[j - 1] j -= 1 i += 1 k += 1 q += 1 # Driver Code a = [1, 3, 5, 7, 2, 4, 6, 8] n = len(a) shuffleArray(a, int(n / 2)) for i in range(0, n): print(a[i], end = " ") # This code is contributed by Smitha Dinesh Semwal.

Output:

1 2 3 4 5 6 7 8

**Time Complexity:** O(n^{2})

**Method 2: (Divide and Conquer)**

The idea is to use Divide and Conquer Technique. Divide the given array into half (say arr1[] and arr2[]) and swap second half element of arr1[] with first half element of arr2[]. Recursively do this for arr1 and arr2.

Let us explain with the help of an example.

- Let the array be a1, a2, a3, a4, b1, b2, b3, b4
- Split the array into two halves: a1, a2, a3, a4 : b1, b2, b3, b4
- Exchange element around the center: exchange a3, a4 with b1, b2 correspondingly.

you get: a1, a2, b1, b2, a3, a4, b3, b4 - Recursively spilt a1, a2, b1, b2 into a1, a2 : b1, b2

then split a3, a4, b3, b4 into a3, a4 : b3, b4. - Exchange elements around the center for each subarray we get:

a1, b1, a2, b2 and a3, b3, a4, b4.

Note: This solution only handles the case when n = 2^{i} where i = 0, 1, 2, …etc.

Below is implementation of this approach:

## C++

// C++ Effective program to shuffle an array of size 2n #include <bits/stdc++.h> using namespace std; // function to shuffle an array of size 2n void shufleArray(int a[], int f, int l) { // If only 2 element, return if (l - f == 1) return; // finding mid to divide the array int mid = (f + l) / 2; // using temp for swapping first half of second array int temp = mid + 1; // mmid is use for swapping second half for first array int mmid = (f + mid) / 2; // Swapping the element for (int i = mmid + 1; i <= mid; i++) swap(a[i], a[temp++]); // Recursively doing for first half and second half shufleArray(a, f, mid); shufleArray(a, mid + 1, l); } // Driven Program int main() { int a[] = { 1, 3, 5, 7, 2, 4, 6, 8 }; int n = sizeof(a) / sizeof(a[0]); shufleArray(a, 0, n - 1); for (int i = 0; i < n; i++) cout << a[i] << " "; return 0; }

## Java

// Java Effective program to shuffle an array of size 2n import java.util.Arrays; public class GFG { // method to shuffle an array of size 2n static void shufleArray(int a[], int f, int l) { // If only 2 element, return if (l - f == 1) return; // finding mid to divide the array int mid = (f + l) / 2; // using temp for swapping first half of second array int temp = mid + 1; // mmid is use for swapping second half for first array int mmid = (f + mid) / 2; // Swapping the element for (int i = mmid + 1; i <= mid; i++) { // swap a[i], a[temp++] int temp1 = a[i]; a[i] = a[temp]; a[temp++] = temp1; } // Recursively doing for first half and second half shufleArray(a, f, mid); shufleArray(a, mid + 1, l); } // Driver Method public static void main(String[] args) { int a[] = { 1, 3, 5, 7, 2, 4, 6, 8 }; shufleArray(a, 0, a.length - 1); System.out.println(Arrays.toString(a)); } }

## Python3

# Python3 effective program to # shuffle an array of size 2n # Function to shuffle an array of size 2n def shufleArray(a, f, l): # If only 2 element, return if (l - f == 1): return # Finding mid to divide the array mid = int((f + l) / 2) # Using temp for swapping first # half of second array temp = mid + 1 # Mid is use for swapping second # half for first array mmid = int((f + mid) / 2) # Swapping the element for i in range(mmid + 1, mid + 1): (a[i], a[temp]) = (a[temp], a[i]) temp += 1 # Recursively doing for first # half and second half shufleArray(a, f, mid) shufleArray(a, mid + 1, l) # Driver Code a = [1, 3, 5, 7, 2, 4, 6, 8] n = len(a) shufleArray(a, 0, n - 1) for i in range(0, n): print(a[i], end = " ") # This code is contributed by Smitha Dinesh Semwal

Output:

1 2 3 4 5 6 7 8

**Time Complexity:** O(n log n)

This article is contributed by **Anuj Chauhan**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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