**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 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:**

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

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.

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