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# Shuffle 2n integers in format {a1, b1, a2, b2, a3, b3, ……, an, bn} without using extra space

• Difficulty Level : Hard
• Last Updated : 11 Aug, 2021

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 ``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);` `    ``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.`

## C#

 `// C# Naive program to shuffle an``// array of size 2n``using` `System;` `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 Code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] a = { 1, 3, 5, 7, 2, 4, 6, 8 };` `        ``shuffleArray(a, a.Length / 2);``        ``for` `(``int` `i = 0; i < a.Length; i++)``            ``Console.Write(a[i] + ``" "``);``    ``}``}` `// This code is contributed``// by ChitraNayal`

## Javascript

 ``

Output:

`1 2 3 4 5 6 7 8 `

Time Complexity: O(n2)

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.

1. Let the array be a1, a2, a3, a4, b1, b2, b3, b4
2. Split the array into two halves: a1, a2, a3, a4 : b1, b2, b3, b4
3. Exchange element around the center: exchange a3, a4 with b1, b2 correspondingly.
you get: a1, a2, b1, b2, a3, a4, b3, b4
4. Recursively split a1, a2, b1, b2 into a1, a2 : b1, b2
then split a3, a4, b3, b4 into a3, a4 : b3, b4.
5. 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 = 2i where i = 0, 1, 2, …etc.

Below is implementation of this approach:

## C++

 `// C++ Effective  program to shuffle an array of size 2n` `#include ``using` `namespace` `std;` `// function to shuffle an array of size 2n``void` `shufleArray(``int` `a[], ``int` `f, ``int` `l)``{``    ``if` `(f > l) {``        ``return``;``    ``}` `    ``// 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);` `    ``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` `(f > l)``            ``return``;` `        ``// 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` `(f > l):``        ``return` `    ``# 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 the 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`

## C#

 `// C# program program to merge two``// sorted arrays with O(1) extra space.``using` `System;` `// method to shuffle an array of size 2n` `public` `class` `GFG {``    ``// method to shuffle an array of size 2n``    ``static` `void` `shufleArray(``int``[] a, ``int` `f, ``int` `l)``    ``{``        ``if` `(f > l)``            ``return``;` `        ``// 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()``    ``{``        ``int``[] a = { 1, 3, 5, 7, 2, 4, 6, 8 };` `        ``shufleArray(a, 0, a.Length - 1);``        ``for` `(``int` `i = 0; i < a.Length; i++)``            ``Console.Write(a[i] + ``" "``);``    ``}``}` `/*This code is contributed by 29AjayKumar*/`

## Javascript

 ``

Output:

`1 2 3 4 5 6 7 8 `

Time Complexity: O(n log n)
Linear time solution

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