Even numbers at even index and odd numbers at odd index

Given an array of size n containing equal number of odd and even numbers. The problem is to arrange the numbers in such a way that all the even numbers get the even index and odd numbers get the odd index. Required auxiliary space is O(1).
Examples :

Input : arr[] = {3, 6, 12, 1, 5, 8}
Output : 6 3 12 1 8 5 

Input : arr[] = {10, 9, 7, 18, 13, 19, 4, 20, 21, 14}
Output : 10 9 18 7 20 19 4 13 14 21 

Source: Amazon Interview Experience | Set 410.

Approach : 

• Start from the left and keep two index one for even position and other for odd positions.
• Traverse these index from left.
• At even position there should be even number and at odd positions, there should be odd number.
• Whenever there is mismatch , we swap the values at odd and even index.

Below is the implementation of the above approach :

CPP

// C++ implementation to arrange
// odd and even numbers
#include <bits/stdc++.h>
using namespace std;

// function to arrange odd and even numbers
void arrangeOddAndEven(int arr[], int n)
{
   int oddInd = 1;
    int evenInd = 0;
    while (true)
    {
        while (evenInd < n && arr[evenInd] % 2 == 0)
            evenInd += 2;

        while (oddInd < n && arr[oddInd] % 2 == 1)
            oddInd += 2;

        if (evenInd < n && oddInd < n)
            swap (arr[evenInd], arr[oddInd]);

        else
            break;
    }
}

// function to print the array
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// Driver program to test above
int main()
{
    int arr[] = { 3, 6, 12, 1, 5, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << "Original Array: ";
    printArray(arr, n);

    arrangeOddAndEven(arr, n);

    cout << "\nModified Array: ";
    printArray(arr, n);

    return 0;
}

Java

// Java implementation to
// arrange odd and even numbers

import java.util.*;
import java.lang.*;

class GfG {

// function to arrange
// odd and even numbers
public static void arrangeOddAndEven(int arr[], int n)
{
    int oddInd = 1;
    int evenInd = 0;
    while (true)
    {
        while (evenInd < n && arr[evenInd] % 2 == 0)
            evenInd += 2;

        while (oddInd < n && arr[oddInd] % 2 == 1)
            oddInd += 2;

        if (evenInd < n && oddInd < n)
            {
                int temp = arr[evenInd];
                arr[evenInd] = arr[oddInd];
                arr[oddInd] = temp;
            }

        else
            break;
    }
}

// function to print the array
public static void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}

// Driver function
public static void main(String argc[]){
    int arr[] = { 3, 6, 12, 1, 5, 8 };
    int n = 6;

    System.out.print("Original Array: ");
    printArray(arr, n);

    arrangeOddAndEven(arr, n);

    System.out.print("\nModified Array: ");
    printArray(arr, n);
}
}

// This code is contributed by Sagar Shukla

Python3

# Python3 implementation to
# arrange odd and even numbers

def arrangeOddAndEven(arr,  n):

    oddInd = 1
    evenInd = 0
    while (True):

        while (evenInd < n and arr[evenInd] % 2 == 0):
            evenInd += 2

        while (oddInd < n and arr[oddInd] % 2 == 1):
            oddInd += 2

        if (evenInd < n and oddInd < n):
                temp = arr[evenInd]
                arr[evenInd] = arr[oddInd]
                arr[oddInd] = temp;

        else:
            break

# function to print the array
def printArray(arr,  n):
    for i in range(0,n):
        print(arr[i] , "",end="")


# Driver function
def main():
    arr = [ 3, 6, 12, 1, 5, 8 ]
    n = 6

    print("Original Array: ",end="")
    printArray(arr, n)

    arrangeOddAndEven(arr, n)

    print("\nModified Array: ",end="")
    printArray(arr, n)

if __name__ == '__main__':
    main()
# This code is contributed by 29AjayKumar

C#

// C# implementation to
// arrange odd and even numbers
using System;

class GFG {

    // function to arrange
    // odd and even numbers
    public static void arrangeOddAndEven(int[] arr, int n)
    {
     int oddInd = 1;
     int evenInd = 0;
    while (true)
    {
        while (evenInd < n && arr[evenInd] % 2 == 0)
            evenInd += 2;

        while (oddInd < n && arr[oddInd] % 2 == 1)
            oddInd += 2;

        if (evenInd < n && oddInd < n)
            {
                int temp = arr[evenInd];
                arr[evenInd] = arr[oddInd];
                arr[oddInd] = temp;
            }

        else
            break;
    }
    }

    // function to print the array
    public static void printArray(int[] arr, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }

    // Driver function
    public static void Main()
    {
        int[] arr = { 3, 6, 12, 1, 5, 8 };
        int n = 6;

        Console.Write("Original Array: ");
        printArray(arr, n);

        arrangeOddAndEven(arr, n);

        Console.Write("\nModified Array: ");
        printArray(arr, n);
    }
}

// This code is contributed by Sam007

Javascript

<script>

// Javascript implementation to arrange
// odd and even numbers

// function to arrange odd and even numbers
function arrangeOddAndEven(arr, n)
{
    let oddInd = 1;
    let evenInd = 0;
    while (true)
    {
        while (evenInd < n && arr[evenInd] % 2 == 0)
            evenInd += 2;

        while (oddInd < n && arr[oddInd] % 2 == 1)
            oddInd += 2;

        if (evenInd < n && oddInd < n)
        {
            let temp;
            temp = arr[evenInd];
            arr[evenInd] = arr[oddInd];
            arr[oddInd] = temp;
        }
        else
            break;
    }
}

// function to print the array
function printArray(arr, n)
{
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}

// Driver program to test above
    let arr = [ 3, 6, 12, 1, 5, 8 ];
    let n = arr.length;

    document.write("Original Array: ");
    printArray(arr, n);

    arrangeOddAndEven(arr, n);

    document.write("<br>" + "Modified Array: ");
    printArray(arr, n);

// This code is contributed by Mayank Tyagi

</script>

Output :

Original Array: 3 6 12 1 5 8 
Modified Array: 6 3 12 1 8 5 

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(1), as we are not using any extra space.

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.