# Rearrange array such that even positioned are greater than odd

Given an array A of n elements, sort the array according to the following relations : , if i is even. , if i is odd.
Print the resultant array.

Examples :

```Input : A[] = {1, 2, 2, 1}
Output :  1 2 1 2
Explanation :
For 1st element, 1  1, i = 2 is even.
3rd element, 1  1, i = 4 is even.

Input : A[] = {1, 3, 2}
Output : 1 3 2
Explanation :
Here, the array is also sorted as per the conditions.
1  1 and 2 < 3.
```

## Recommended: Please solve it on PRACTICE first, before moving on to the solution.

Method 1 –

Observe that array consists of [n/2] even positioned elements. If we assign largest [n/2] elements to the even positions and rest of the elements to the odd positions, our problem is solved. Because element at odd position will always be less than the element at even position as it is maximum element and vice versa. Sort the array and assign the first [n/2] elements at even positions.

Below is the implementation of above approach:

## C++

 `// C++ program to rearrange the elements ` `// in array such that even positioned are ` `// greater than odd positioned elements ` `#include ` `using` `namespace` `std; ` ` `  `void` `assign(``int` `a[], ``int` `n) ` `{ ` `    ``// Sort the array ` `    ``sort(a, a + n); ` ` `  `    ``int` `ans[n]; ` `    ``int` `p = 0, q = n - 1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// Assign even indexes with maximum elements ` `        ``if` `((i + 1) % 2 == 0) ` `            ``ans[i] = a[q--]; ` ` `  `        ``// Assign odd indexes with remaining elements ` `        ``else` `            ``ans[i] = a[p++]; ` `    ``} ` ` `  `    ``// Print result ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << ans[i] << ``" "``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `A[] = { 1, 3, 2, 2, 5 }; ` `    ``int` `n = ``sizeof``(A) / ``sizeof``(A); ` `    ``assign(A, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to rearrange the elements ` `// in array such that even positioned are ` `// greater than odd positioned elements ` `import` `java.io.*; ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``static` `void` `assign(``int` `a[], ``int` `n) ` `    ``{ ` ` `  `        ``// Sort the array ` `        ``Arrays.sort(a); ` ` `  `        ``int` `ans[] = ``new` `int``[n]; ` `        ``int` `p = ``0``, q = n - ``1``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `            ``// Assign even indexes with maximum elements ` `            ``if` `((i + ``1``) % ``2` `== ``0``) ` `                ``ans[i] = a[q--]; ` ` `  `            ``// Assign odd indexes with remaining elements ` `            ``else` `                ``ans[i] = a[p++]; ` `        ``} ` ` `  `        ``// Print result ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``System.out.print(ans[i] + ``" "``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `A[] = { ``1``, ``3``, ``2``, ``2``, ``5` `}; ` `        ``int` `n = A.length; ` `        ``assign(A, n); ` `    ``} ` `} ` ` `  `// This code is contributed by Nikita Tiwari. `

## Python3

 `# Python3 code to rearrange the ` `# elements in array such that ` `# even positioned are greater ` `# than odd positioned elements ` ` `  `def` `assign(a, n): ` `     `  `    ``# Sort the array ` `    ``a.sort() ` `     `  `    ``ans ``=` `[``0``] ``*` `n ` `    ``p ``=` `0` `    ``q ``=` `n ``-` `1` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Assign even indexes with ` `        ``# maximum elements ` `        ``if` `(i ``+` `1``) ``%` `2` `=``=` `0``: ` `            ``ans[i] ``=` `a[q] ` `            ``q ``=` `q ``-` `1` `         `  `        ``# Assign odd indexes with ` `        ``# remaining elements ` `        ``else``: ` `            ``ans[i] ``=` `a[p] ` `            ``p ``=` `p ``+` `1` `             `  `    ``# Print result ` `    ``for` `i ``in` `range``(n): ` `        ``print``(ans[i], end ``=` `" "``) ` ` `  `# Driver Code ` `A ``=` `[ ``1``, ``3``, ``2``, ``2``, ``5` `] ` `n ``=` `len``(A) ` `assign(A, n) ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

## C#

 `// C# program to rearrange the elements ` `// in array such that even positioned are ` `// greater than odd positioned elements ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``static` `void` `assign(``int``[] a, ``int` `n) ` `    ``{ ` `        ``// Sort the array ` `        ``Array.Sort(a); ` ` `  `        ``int``[] ans = ``new` `int``[n]; ` `        ``int` `p = 0, q = n - 1; ` `        ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `            ``// Assign even indexes with maximum elements ` `            ``if` `((i + 1) % 2 == 0) ` `                ``ans[i] = a[q--]; ` ` `  `            ``// Assign odd indexes with remaining elements ` `            ``else` `                ``ans[i] = a[p++]; ` `        ``} ` ` `  `        ``// Print result ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``Console.Write(ans[i] + ``" "``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] A = { 1, 3, 2, 2, 5 }; ` `        ``int` `n = A.Length; ` `        ``assign(A, n); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```  1 5 2 3 2
```

Method 2 –
One other approach is to traverse the array from the second element and swap the element with the previous one if the condition is not satisfied. This is implemented as follows:

## C++

 `// C++ program to rearrange the elements ` `// in the array such that even positioned are ` `// greater than odd positioned elements ` `#include ` `using` `namespace` `std; ` ` `  `// swap two elements ` `void` `swap(``int``* a, ``int``* b) ` `{ ` `    ``int` `temp = *a; ` `    ``*a = *b; ` `    ``*b = temp; ` `} ` ` `  `void` `rearrange(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 1; i < n; i++) { ` `        ``// if index is even ` `        ``if` `(i % 2 == 0) { ` `            ``if` `(arr[i] > arr[i - 1]) ` `                ``swap(&arr[i - 1], &arr[i]); ` `        ``} ` `        ``// if index is odd ` `        ``else` `{ ` `            ``if` `(arr[i] < arr[i - 1]) ` `                ``swap(&arr[i - 1], &arr[i]); ` `        ``} ` `    ``} ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `n = 5; ` `    ``int` `arr[] = { 1, 3, 2, 2, 5 }; ` `    ``rearrange(arr, n); ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << arr[i] << ``" "``; ` `    ``cout << ``"\n"``; ` `    ``return` `0; ` `} `

Output:

```  1 3 2 5 2
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : jit_t, mohdmohtashimnawaz

Article Tags :
Practice Tags :

11

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.