# Divide and Conquer - Rearrange array elements in special order

Objec­tive:  Given an array of integers of size 2n, write an algorithm to arrange them such that first n elements and last n elements are set up in alternative manner. Say n = 3 and 2n elements are {x1, x2, x3, y1, y2, y3} , then result should be {x1, y1, x2, y2, x3, y3}

Example:

```A [] = {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}, n= 5
Output: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}```

Approaches:

Brute Force:

One by one shift all the elements from the second half of the array to their correct positions in the left half of the array.

Time Complexity: O(n^2)

See the explanation below-

Code:

Output:

`1 2 3 4 5 6 7 8 9 10`

Divide and Conquer:

1. This solution will work only if the total number of elements is in 2i So total elements are either 2 or 4 or 8 or 16 ….and so on.
2. Total length is 2n, take n elements around middle element.
3. Swap n/2 elements on the left side from the middle element with n/2 elements on the right side from the middle element.
4. Now divide the array into 2 parts, first n elements and last n elements.
5. Repeat step 2, 3 on both the parts recursively.

Time Complexity: O(nlogn)

See the explanation below-

Code:

Output:

`[1, 2, 3, 4, 5, 6, 7, 8]`