Segregate Even and Odd numbers

1.6

Given an array A[], write a function that segregates even and odd numbers. The functions should put all even numbers first, and then odd numbers.

Example

Input  = {12, 34, 45, 9, 8, 90, 3}
Output = {12, 34, 8, 90, 45, 9, 3} 

In the output, order of numbers can be changed, i.e., in the above example 34 can come before 12 and 3 can come before 9.

The problem is very similar to our old post Segregate 0s and 1s in an array, and both of these problems are variation of famous Dutch national flag problem.


Algorithm: segregateEvenOdd()
1) Initialize two index variables left and right:  
            left = 0,  right = size -1 
2) Keep incrementing left index until we see an odd number.
3) Keep decrementing right index until we see an even number.
4) If lef < right then swap arr[left] and arr[right]

Implementation:

C/C++

// C program to segregate even and odd elements of array
#include<stdio.h>

/* Function to swap *a and *b */
void swap(int *a, int *b);

void segregateEvenOdd(int arr[], int size)
{
    /* Initialize left and right indexes */
    int left = 0, right = size-1;
    while (left < right)
    {
        /* Increment left index while we see 0 at left */
        while (arr[left]%2 == 0 && left < right)
            left++;

        /* Decrement right index while we see 1 at right */
        while (arr[right]%2 == 1 && left < right)
            right--;

        if (left < right)
        {
            /* Swap arr[left] and arr[right]*/
            swap(&arr[left], &arr[right]);
            left++;
            right--;
        }
    }
}

/* UTILITY FUNCTIONS */
void swap(int *a, int *b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}

/* driver program to test */
int main()
{
    int arr[] = {12, 34, 45, 9, 8, 90, 3};
    int arr_size = sizeof(arr)/sizeof(arr[0]);
    int i = 0;

    segregateEvenOdd(arr, arr_size);

    printf("Array after segregation ");
    for (i = 0; i < arr_size; i++)
        printf("%d ", arr[i]);

    return 0;
}

Java

// Java program to segregate even and odd elements of array
import java.io.*;

class SegregateOddEven
{
    static void segregateEvenOdd(int arr[])
    {
        /* Initialize left and right indexes */
        int left = 0, right = arr.length - 1;
        while (left < right)
        {
            /* Increment left index while we see 0 at left */
            while (arr[left]%2 == 0 && left < right)
                left++;

            /* Decrement right index while we see 1 at right */
            while (arr[right]%2 == 1 && left < right)
                right--;

            if (left < right)
            {
                /* Swap arr[left] and arr[right]*/
                int temp = arr[left];
                arr[left] = arr[right];
                arr[right] = temp;
                left++;
                right--;
            }
        }
    }

    /* Driver program to test above functions */
    public static void main (String[] args)
    {
        int arr[] = {12, 34, 45, 9, 8, 90, 3};

        segregateEvenOdd(arr);

        System.out.print("Array after segregation ");
        for (int i = 0; i < arr.length; i++)
            System.out.print(arr[i]+" ");
    }
}
/*This code is contributed by Devesh Agrawal*/

Python

# Python program to segregate even and odd elements of array

def segregateEvenOdd(arr):

    # Initialize left and right indexes
    left,right = 0,len(arr)-1

    while left < right:

        # Increment left index while we see 0 at left
        while (arr[left]%2==0 and left < right):
            left += 1

        # Decrement right index while we see 1 at right
        while (arr[right]%2 == 1 and left < right):
            right -= 1

        if (left < right):
              # Swap arr[left] and arr[right]*/
              arr[left],arr[right] = arr[right],arr[left]
              left += 1
              right = right-1


# Driver function to test above function
arr = [12, 34, 45, 9, 8, 90, 3]
segregateEvenOdd(arr)
print ("Array after segregation "),
for i in range(0,len(arr)):
    print arr[i],
# This code is contributed by Devesh Agrawal


Output:
Array after segregation 12 34 90 8 9 45 3 

Time Complexity: O(n)

Alternate Implementation (Lomuto partition):

C++

// A Lomuto partition based scheme to segregate
// even and odd numbers.
#include <iostream>
using namespace std;

// fuction is rearrange the array in given way.
void rearrangeEvenAndOdd(int arr[], int n)
{
    // variables
    int j = -1;

    // quick sort method
    for (int i = 0; i < n; i++) {

        // if array of element
        // is odd then swap
        if (arr[i] % 2 == 0) {

            // increment j by one
            j++;

            // swap the element
            swap(arr[i], arr[j]);
        }
    }
}

int main()
{
    int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
    int n = sizeof(arr) / sizeof(arr[0]);

    rearrangeEvenAndOdd(arr, n);

    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
// This code is contributed by devagarwalmnnit

Java

// A Lomuto partition based scheme to segregate
// even and odd numbers.
import java.io.*;

class GFG 
{
    // fuction is rearrange the array in given way.
    static void rearrangeEvenAndOdd(int arr[], int n)
    {
        // variables
        int j = -1,temp;
    
        // quick sort method
        for (int i = 0; i < n; i++) {
    
            // if array of element
            // is odd then swap
            if (arr[i] % 2 == 0) {
    
                // increment j by one
                j++;
    
                // swap the element
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    }
    
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
        int n = arr.length;
    
        rearrangeEvenAndOdd(arr, n);
    
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
}

// This code is contributed by Nikita Tiwari.

Python3

# A Lomuto partition based scheme to 
# segregate even and odd numbers.

# fuction is rearrange the 
# array in given way.
def rearrangeEvenAndOdd(arr, n) :
    # variables
    j = -1

    # quick sort method
    for i in range(0, n) :
        
        # if array of element
        # is odd then swap
        if (arr[i] % 2 == 0) :
            # increment j by one
            j = j + 1

            # swap the element
            temp = arr[i]
            arr[i] = arr[j]
            arr[j] = temp
        

# Driver code        
arr = [ 12, 10, 9, 45, 2, 10, 10, 45 ]
n = len(arr)

rearrangeEvenAndOdd(arr, n)

for i in range(0,n) :
    print( arr[i] ,end= " ")
    
        
# This code is contributed by Nikita Tiwari.


Output:

12 10 2 10 10 45 9 45

Time Complexity: O(n)

References:
http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Sort/Flag/

Asked in: MakeMyTrip,Paytm

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