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Segregate Even and Odd numbers

  • Difficulty Level : Easy
  • Last Updated : 04 Jun, 2021

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:  

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

In the output, the 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 even number.
3) Keep decrementing right index until we see an odd number.
4) If left < right then swap arr[left] and arr[right]

Implementation:  

C++




// C++ program to segregate even and odd elements of array
#include <iostream>
using namespace std;
 
/* 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 code */
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);
 
    cout <<"Array after segregation ";
    for (i = 0; i < arr_size; i++)
        cout << arr[i] << " ";
 
    return 0;
}
 
// This code is contributed by shubhamsingh10


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


C#




// C# program to segregate even
// and odd elements of array
using System;
 
class GFG
{
    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 ()
    {
        int []arr = {12, 34, 45, 9, 8, 90, 3};
 
        segregateEvenOdd(arr);
 
        Console.Write("Array after segregation ");
        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i]+" ");
    }
}
 
//This code is contributed by Sam007


PHP




<?php
// PHP program to segregate even and
// odd elements of array
 
function segregateEvenOdd(&$arr, $size)
{
    // Initialize left and right indexes
    $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
function swap(&$a, &$b)
{
    $temp = $a;
    $a = $b;
    $b = $temp;
}
 
// Driver Code
$arr = array(12, 34, 45, 9, 8, 90, 3);
$arr_size = count($arr);
 
segregateEvenOdd($arr, $arr_size);
 
echo "Array after segregation ";
for ($i = 0; $i < $arr_size; $i++)
    echo $arr[$i]." ";
 
// This code is contributed
// by rathbhupendra
?>


Javascript




<script>
 
// javascript program to segregate even
// and odd elements of array
 
 
function segregateEvenOdd(arr)
{
    /* Initialize left and right indexes */
       var 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]*/
            var temp = arr[left];
            arr[left] = arr[right];
            arr[right] = temp;
            left++;
            right--;
        }
    }
}
 
/* Driver program to test above functions */
   var arr = [12, 34, 45, 9, 8, 90, 3];
 
   segregateEvenOdd(arr);
 
   document.write("Array after segregation ");
   for (i = 0; i < arr.length; i++)
       document.write(arr[i]+" ");
 
// This code is contributed by 29AjayKumar
 
</script>


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;
 
// Function to 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
{
    // function to 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.
 
# function to 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.


C#




// A Lomuto partition based scheme
// to segregate even and odd numbers.
using System;
 
class GFG
{
    // function to 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()
    {
        int []arr = { 12, 10, 9, 45, 2,
                            10, 10, 45 };
        int n = arr.Length;
     
        rearrangeEvenAndOdd(arr, n);
     
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
}
 
// This code is contributed by Sam007


PHP




<?php
// A Lomuto partition based scheme to
// segregate even and odd numbers.
 
// function to rearrange the array
// in given way.
function rearrangeEvenAndOdd(&$arr, $n)
{
    // variables
    $j = -1; $temp;
 
    // quick sort method
    for ($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
$arr = array( 12, 10, 9, 45, 2, 10, 10, 45 );
$n = sizeof($arr);
 
rearrangeEvenAndOdd($arr, $n);
 
for ($i = 0; $i < $n; $i++)
    echo($arr[$i] . " ");
 
// This code is contributed by Code_Mech.


Javascript




<script>
 
// A Lomuto partition based scheme to segregate
// even and odd numbers.
 
// Function to rearrange the array in given way.
function rearrangeEvenAndOdd(arr, n)
{
     
    // Variables
    var j = -1, temp;
 
    // Quick sort method
    for(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
var arr = [ 12, 10, 9, 45, 2, 10, 10, 45 ];
var n = arr.length;
 
rearrangeEvenAndOdd(arr, n);
 
for(i = 0; i < n; i++)
    document.write(arr[i] + " ");
 
// This code is contributed by gauravrajput1
 
</script>


Output

12 10 2 10 10 45 9 45 

Time Complexity: O(n)

Alternate Implementation (Using stable partition): 



To implement the above problem, we will use stable_partition in C++. The stable_partition() algorithm arranges the sequence defined by start and end such that all elements for which the predicate specified by pfn returns true come before those for which the predicate returns false. The partitioning is stable. This means that the relative ordering of the sequence is preserved.

Syntax:

template  
BiIter stable_partition(BiIter start, BiIter end, UnPred pfn);

Parameters:

start: the range of elements to reorder
end: the range of elements to reorder
pfn: User-defined predicate function object that defines 
the condition to be satisfied if an element
is to be classified. A predicate takes single argument 
and returns true or false.
Return Value:
Returns an iterator to the beginning of the elements 
for which the predicate is false.

This function attempts to allocate a temporary buffer. If the allocation fails, the less efficient algorithm is chosen.

Below is the implementation of the above logic.

Code:

C++14




// CPP program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to rearrange the array in given way.
void rearrangeEvenAndOdd(vector<int>v)
{
     
    // Using stable partition with lambda expression
    stable_partition(v.begin(), v.end(),
                     [](auto a) { return a % 2 == 0; });
 
    for (int num : v)
        cout << num << " ";
}
 
// Driver Code
int main()
{
    vector<int> v = { 12, 10, 9, 45, 2, 10, 10, 45 };
     
    // Function Call
    rearrangeEvenAndOdd(v);
    return 0;
}
// This code is contributed by Chirag Shilwant


Output

12 10 2 10 10 9 45 45 

Time Complexity:

If enough extra memory is available, linear in the distance between first and last i.e (O(N) ,where N is the distance between first and last). It applies predicate (i.e 3rd parameter of above code) exactly once to each element, and performs up to that many element moves.

Otherwise, performs up to N*log(N) element swaps (where N is the distance above). It also applies predicate exactly once to each element.

Alternate Implementation:

  1. Using two pointers i and j , i will point index 0 and j will point the last index.
  2. Run a while loop; if a[i] is odd and a[j] is even then we will swap them else we will decrement j.

Code:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to segregate
void segregate(int a[], int n)
{
    int i = 0, j = (n - 1);
 
    // Iterate while j >= i
    while (j >= i) {
       
        // Check is a[i] is even
        // or odd
        if (a[i] % 2 != 0)
        {
            if (a[j] % 2 == 0)
            {
                
                // Swap a[i] and a[j]
                swap(a[i], a[j]);
                i++;
                j--;
            }
            else
                j--;
        }
        else
            i++;
    }
 
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
}
 
// Driver Code
int main()
{
    int a[] = { 1,2,3,4,5,6 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // Function Call
    segregate(a, n);
    return 0;
}


Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to segregate
static void segregate(int a[], int n)
{
    int i = 0, j = (n - 1);
 
    // Iterate while j >= i
    while (j >= i)
    {
         
        // Check is a[i] is even
        // or odd
        if (a[i] % 2 != 0)
        {
            if (a[j] % 2 == 0)
            {
                
                // Swap a[i] and a[j]
                a = swap(a, i, j);
                i++;
                j--;
            }
            else
                j--;
        }
        else
            i++;
    }
 
    for(i = 0; i < n; i++)
        System.out.print(a[i] + " ");
}
 
static int[] swap(int []arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
 
// Driver Code
public static void main(String[] args)
{
    int a[] = { 1, 2, 3, 4, 5, 6 };
    int n = a.length;
 
    // Function Call
    segregate(a, n);
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python3 program for the above approach
 
# Function to segregate
def segregate(a, n):
     
    i = 0
    j = (n - 1)
  
    # Iterate while j >= i
    while (j >= i):
     
        # Check is a[i] is even
        # or odd
        if (a[i] % 2 != 0):
            if (a[j] % 2 == 0):
                 
                # Swap a[i] and a[j]
                a[i], a[j] = a[j], a[i]
                i += 1
                j -= 1
             
            else:
                j -= 1
        else:
            i += 1;
             
    for i in range(n):
        print(a[i], end = " ")
 
# Driver Code
a = [ 1, 2, 3, 4, 5, 6 ]
n = len(a)
 
segregate(a, n)
 
#  This code is contributed by rag2127


C#




// C# program for the above approach
using System;
class GFG
{
 
    // Function to segregate
    static void segregate(int[] a, int n)
    {
        int i = 0, j = (n - 1);
 
        // Iterate while j >= i
        while (j >= i)
        {
 
            // Check is a[i] is even
            // or odd
            if (a[i] % 2 != 0) {
                if (a[j] % 2 == 0) {
 
                    // Swap a[i] and a[j]
                    int temp = a[i];
                    a[i] = a[j];
                    a[j] = temp;
                    i++;
                    j--;
                }
                else
                    j--;
            }
            else
                i++;
        }
 
        for (i = 0; i < n; i++)
            Console.Write(a[i] + " ");
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int[] a = { 1, 2, 3, 4, 5, 6 };
        int n = a.Length;
 
        // Function Call
        segregate(a, n);
    }
}
 
// This code is contributed by ukasp.


Javascript




// JAVASCRIPT program for the above approach
 
import java.util.*;
class GFG{
 
// Function to segregate
static void segregate(int a[], int n)
{
    int i = 0, j = (n - 1);
 
    // Iterate while j >= i
    while (j >= i) {
       
        // Check is a[i] is even
        // or odd
        if (a[i] % 2 != 0)
        {
            if (a[j] % 2 == 0)
            {
                
                // Swap a[i] and a[j]
                a = swap(a,i,j);
                i++;
                j--;
            }
            else
                j--;
        }
        else
            i++;
    }
 
    for (i = 0; i < n; i++)
        System.out.print(a[i]+ " ");
}
static int[] swap(int []arr, int i, int j){
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
// Driver Code
public static void main(String[] args)
{
    int a[] = { 1,2,3,4,5,6 };
    int n = a.length;
 
    // Function Call
    segregate(a, n);
}
}
 
// This code contributed by Princi Singh


Output

6 2 4 3 5 1 

Time Complexity: O(N) 

Space Complexity: O(1)

References: 
http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Sort/Flag/
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.




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