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,  ∀ 1 <= i < n
• , if i is odd ,  ∀ 1 <= i < n

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.

Note: Examples are based upon 1-based indexing

Method 1:

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

Note: This solution is considering 1-based indexing

Below is the implementation of the above approach: 

C++

// C++ program to rearrange the elements in array such that
// even positioned are greater than odd positioned elements
#include <bits/stdc++.h>
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[0]);
    assign(A, n);
    return 0;
}

C

// C program to rearrange the elements in array such that
// even positioned are greater than odd positioned elements
#include <stdio.h>
#include <stdlib.h>

// Compare function for qsort
int cmpfunc(const void* a, const void* b)
{
    return (*(int*)a - *(int*)b);
}

void assign(int a[], int n)
{
    // Sort the array
    qsort(a, n, sizeof(int), cmpfunc);

    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++)
        printf("%d ", ans[i]);
}

// Driver Code
int main()
{
    int A[] = { 1, 3, 2, 2, 5 };
    int n = sizeof(A) / sizeof(A[0]);
    assign(A, n);
    return 0;
}

// This code is contributed by Sania Kumari Gupta

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.

Javascript

<script>

// JavaScript program to rearrange the elements
// in array such that even positioned are
// greater than odd positioned elements

    function assign(a, n)
    {

        // Sort the array
        a.sort();

        let ans = [];
        let p = 0, q = n - 1;
        for (let 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 (let i = 0; i < n; i++)
            document.write(ans[i] + " ");
    }

// Driver code

        let A = [ 1, 3, 2, 2, 5 ];
        let n = A.length;
        assign(A, n);

</script>

PHP

<?php
// PHP program to rearrange
// the elements in array such
// that even positioned are
// greater than odd positioned
// elements

function assign($a, $n)
{

    // Sort the array
    sort($a);

    $p = 0; $q = $n - 1;
    for ($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 ($i = 0; $i < $n; $i++)
        echo($ans[$i] . " ");
}

// Driver Code
$A = array( 1, 3, 2, 2, 5 );
$n = sizeof($A);
assign($A, $n);

// This code is contributed by Ajit.
?>

Output

1 5 2 3 2 

Time Complexity: O(n * log n)
Auxiliary Space: O(n), since n extra space has been taken.

Method 2:

One other approach is to traverse the array from the first element till n-1 and swap the element with the next one if the condition is not satisfied. This is implemented as follows: 

Note: This solution is considering 0-based indexing

C++

// C++ program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
#include <bits/stdc++.h>
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 = 0; i < n-1; i+=2) {
            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;
}

Java

// Java program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
import java.io.*;

class GFG {
    public static 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 two elements
                    int temp = arr[i];
                    arr[i] = arr[i - 1];
                    arr[i - 1] = temp;
                }
            }

            // if index is odd
            else {
                if (arr[i] > arr[i - 1]) {

                    // swap two elements
                    int temp = arr[i];
                    arr[i] = arr[i - 1];
                    arr[i - 1] = temp;
                }
            }
        }

        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 5;
        int arr[] = { 1, 3, 2, 2, 5 };
        rearrange(arr, n);
    }
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program to rearrange
# the elements in the array
# such that even positioned are
# greater than odd positioned elements
def rearrange(arr, n):

    for i in range (1, n):

        # if index is even
        if (i % 2 == 0):
            if (arr[i] < arr[i - 1]):
                arr[i - 1], arr[i] = arr[i], arr[i - 1]

        # if index is odd
        else:
            if (arr[i] > arr[i - 1]):
                arr[i - 1], arr[i] = arr[i] , arr[i - 1]

if __name__ == "__main__":
    n = 5
    arr = [1, 3, 2, 2, 5]
    rearrange(arr, n);
    for i in range (n):
        print (arr[i], end = " ")
    print ()

# This code is contributed by Chitranayal

C#

// C# program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
using System;
class GFG
{
public static 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 two elements
          int temp = arr[i];
          arr[i] = arr[i - 1];
          arr[i - 1] = temp;
          }
      }

      // if index is odd
      else
      {
          if (arr[i] > arr[i - 1])
          {

            // swap two elements
            int temp = arr[i];
            arr[i] = arr[i - 1];
            arr[i - 1] = temp;
          }
      }
    }

  for (int i = 0; i < n; i++)
  {
    Console.Write(arr[i] + " ");
  }
}

// Driver code
public static void Main(String []args)
{
    int n = 5;
    int []arr = {1, 3, 2, 2, 5};
    rearrange(arr, n);
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

// JavaScript program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements

  function rearrange(arr, n)
  {
    for(let i = 1; i < n; i++)
    {

      // if index is even
      if(i % 2 == 0)
      {
        if(arr[i] < arr[i - 1])
        {

          // swap two elements
          let temp = arr[i];
          arr[i] = arr[i - 1];
          arr[i - 1] = temp;
        }
      }

      // if index is odd
      else
      {
        if (arr[i] > arr[i - 1])
        {

          // swap two elements
          let temp = arr[i];
          arr[i] = arr[i - 1];
          arr[i - 1] = temp;
        }
      }
    }

    for (let i = 0; i < n; i++)
    {
      document.write(arr[i] + " ");
    }
  }


// Driver code

    let n = 5;
    let arr = [1, 3, 2, 2, 5];
    rearrange(arr, n);

</script>

Output

3 1 2 2 5 

Time Complexity: O(n)
Auxiliary Space: O(1)