# Rearrange array such that even index elements are smaller and odd index elements are greater

Hello Everyone,

Given an array, rearrange the array such that :

1. If index i is even, arr[i] <= arr[i+1]
2. If index i is odd, arr[i] >= arr[i+1]

Note : There can be multiple answers.
Examples:

Input : arr[] = {2, 3, 4, 5}
Output : arr[] = {2, 4, 3, 5}
Explanation : Elements at even indexes are smaller and elements at odd indexes are greater than their next elements

Note : Another valid answer is arr[] = {3, 4, 2, 5} Input :arr[] = {6, 4, 2, 1, 8, 3} Output :arr[] = {4, 6, 1, 8, 2, 3}

This problem is similar to sort an array in wave form.
A simple solution is to sort the array in decreasing order, then starting from second element, swap the adjacent elements.
An efficient solution is to iterate over the array and swap the elements as per the given condition.
If we have an array of length n, then we iterate from index 0 to n-2 and check the given condition.
At any point of time if i is even and arr[i] > arr[i+1], then we swap arr[i] and arr[i+1]. Similarly, if i is odd and
arr[i] < arr[i+1], then we swap arr[i] and arr[i+1].
For the given example:
Before rearrange, arr[] = {2, 3, 4, 5}
Start iterating over the array till index 2 (as n = 4)

First Step:
At i = 0, arr[i] = 2 and arr[i+1] = 3. As i is even and arr[i] < arr[i+1], don’t need to swap.
Second step:
At i = 1, arr[i] = 3 and arr[i+1] = 4. As i is odd and arr[i] < arr[i+1], swap them.
Now arr[] = {2, 4, 3, 5}
Third step:
At i = 2, arr[i] = 3 and arr[i+1] = 5. So, don’t need to swap them

After rearrange, arr[] = {2, 4, 3, 5}

`// CPP code to rearrange an array such that`

`// even index elements are smaller and odd`

`// index elements are greater than their`

`// next.`

`#include <iostream>`

`using` `namespace` `std;`

`void` `rearrange(` `int` `* arr, ` `int` `n)`

`{`

` ` `for` `(` `int` `i = 0; i < n - 1; i++) {`

` ` `if` `(i % 2 == 0 && arr[i] > arr[i + 1])`

` ` `swap(arr[i], arr[i + 1]);`

` ` `if` `(i % 2 != 0 && arr[i] < arr[i + 1])`

` ` `swap(arr[i], arr[i + 1]);`

` ` `}`

`}`

`/* Utility that prints out an array in`

` ` `a line */`

`void` `printArray(` `int` `arr[], ` `int` `size)`

`{`

` ` `for` `(` `int` `i = 0; i < size; i++)`

` ` `cout << arr[i] << ` `" "` `;`

` ` `cout << endl;`

`}`

`/* Driver function to test above functions */`

`int` `main()`

`{`

` ` `int` `arr[] = { 6, 4, 2, 1, 8, 3 };`

` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr);`

` ` `cout << ` `"Before rearranging: \n"` `;`

` ` `printArray(arr, n);`

` ` `rearrange(arr, n);`

` ` `cout << ` `"After rearranging: \n"` `;`

` ` `printArray(arr, n);`

` ` `return` `0;`

`}`

Output:

Before rearranging: 6 4 2 1 8 3 After rearranging: 4 6 1 8 2 3

Time Complexity : O(n)

Auxiliary Space: O(1)