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Segregate even and odd numbers | Set 3

  • Difficulty Level : Easy
  • Last Updated : 23 Jul, 2021

Given an array of integers, segregate even and odd numbers in the array. All the even numbers should be present first, and then the odd numbers.

Examples:  

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Input: arr[] = 1 9 5 3 2 6 7 11
Output: 2 6 5 3 1 9 7 11



Input: arr[] = 1 3 2 4 7 6 9 10
Output:  2 4 6 10 7 1 9 3

We have discussed two different approaches in below posts: 

  1. Segregate Even and Odd numbers
  2. Segregate even and odd numbers | Set 2

Brute-Force Solution

As we need to maintain the order of elements then this can be done in the following steps :

  1. Create a temporary array A of size n and an integer ind which will keep the index of elements inserted .
  2. Initialize ind with zero and iterate over the original array and if even number is found then put that number at A[ind] and then increment the value of ind .
  3. Again iterate over array and if an odd number is found then put it in A[ind] and then increment the value of ind.
  4. Iterate over the temporary array A and copy its values in the original array.

C++




// C++ Implementation of the above approach
#include <iostream>
using namespace std;
void arrayEvenAndOdd(int arr[], int n)
{
    int a[n], ind = 0;
    for (int i = 0; i < n; i++)
    {
        if (arr[i] % 2 == 0)
        {
            a[ind] = arr[i];
            ind++;
        }
    }
    for (int i = 0; i < n; i++)
    {
        if (arr[i] % 2 != 0)
        {
            a[ind] = arr[i];
            ind++;
        }
    }
    for (int i = 0; i < n; i++)
    {
        cout << a[i] << " ";
    }
    cout << endl;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, 2, 4, 7, 6, 9, 10 };
    int n = sizeof(arr) / sizeof(int);
 
    // Function call
    arrayEvenAndOdd(arr, n);
    return 0;
}


Java




// Java Implementation of the above approach
import java.io.*;
class GFG
{
public static void arrayEvenAndOdd(int arr[], int n)
{
    int[] a;
    a = new int[n];
    int ind = 0;
    for (int i = 0; i < n; i++)
    {
        if (arr[i] % 2 == 0)
        {
            a[ind] = arr[i];
            ind++;
        }
    }
    for (int i = 0; i < n; i++)
    {
        if (arr[i] % 2 != 0)
        {
            a[ind] = arr[i];
            ind++;
        }
    }
    for (int i = 0; i < n; i++)
    {
        System.out.print(a[i] + " ");
    }
    System.out.println("");
}
    // Driver code
    public static void main (String[] args)
    {
       
      int arr[] = { 1, 3, 2, 4, 7, 6, 9, 10 };
            int n = arr.length;
 
            // Function call
            arrayEvenAndOdd(arr, n);
    }
}
 
// This code is contributed by rohitsingh07052.


Python3




# Python3 implementation of the above approach
def arrayEvenAndOdd(arr,  n):
     
    ind = 0;
    a = [0 for i in range(n)]
     
    for i in range(n):
        if (arr[i] % 2 == 0):
            a[ind] = arr[i]
            ind += 1
 
    for i in range(n):
        if (arr[i] % 2 != 0):
            a[ind] = arr[i]
            ind += 1
 
    for i in range(n):
        print(a[i], end = " ")
         
    print()
 
# Driver code
arr = [ 1, 3, 2, 4, 7, 6, 9, 10 ]
n = len(arr)
 
# Function call
arrayEvenAndOdd(arr, n)
 
# This code is contributed by rohitsingh07052


C#




// C# implementation of the above approach
using System;
 
class GFG{
     
static void arrayEvenAndOdd(int[] arr, int n)
{
    int[] a = new int[n];
    int ind = 0;
     
    for(int i = 0; i < n; i++)
    {
        if (arr[i] % 2 == 0)
        {
            a[ind] = arr[i];
            ind++;
        }
    }
    for(int i = 0; i < n; i++)
    {
        if (arr[i] % 2 != 0)
        {
            a[ind] = arr[i];
            ind++;
        }
    }
    for(int i = 0; i < n; i++)
    {
        Console.Write(a[i] + " ");
    }
}
 
// Driver code
static public void Main()
{
    int[] arr = { 1, 3, 2, 4, 7, 6, 9, 10 };
    int n = arr.Length;
     
    arrayEvenAndOdd(arr, n);
}
}
 
// This code is contributed by Potta Lokesh


Javascript




<script>
 
// JavaScript Implementation of the above approach
 
function arrayEvenAndOdd(arr, n){
    let a= [];
    let ind = 0;
    for (let i = 0; i < n; i++)
    {
        if (arr[i] % 2 == 0)
        {
            a[ind] = arr[i];
            ind++;
        }
    }
    for (let i = 0; i < n; i++)
    {
        if (arr[i] % 2 != 0)
        {
            a[ind] = arr[i];
            ind++;
        }
    }
    for (let i = 0; i < n; i++)
    {
        document.write(a[i] +" ");
    }
    document.write('\n');
}
 
// Driver code
 
let arr = [ 1, 3, 2, 4, 7, 6, 9, 10 ];
let n = arr.length;
 
// Function call
arrayEvenAndOdd(arr, n);
 
</script>


Output

2 4 6 10 1 3 7 9 

Time complexity: O(n)
Auxiliary space: O(n) 

Efficient Approach:

The optimization for above approach is based on Lomuto‚Äôs Partition Scheme 

  1. Maintain a pointer to the position before first odd element in the array.
  2. Traverse the array and if even number is encountered then swap it with the first odd element.
  3. Continue the traversal.

Below is the implementation of the above approach: 

C++




// CPP code to segregate even odd
// numbers in an array
#include <bits/stdc++.h>
using namespace std;
 
// Function to segregate even odd numbers
void arrayEvenAndOdd(int arr[], int n)
{
 
    int i = -1, j = 0;
    int t;
    while (j != n) {
        if (arr[j] % 2 == 0) {
            i++;
 
            // Swapping even and odd numbers
            swap(arr[i], arr[j]);
        }
        j++;
    }
 
    // Printing segregated array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, 2, 4, 7, 6, 9, 10 };
    int n = sizeof(arr) / sizeof(int);
    arrayEvenAndOdd(arr, n);
    return 0;
}


Java




// java code to segregate even odd
// numbers in an array
public class GFG {
      
    // Function to segregate even
    // odd numbers
    static void arrayEvenAndOdd(
                  int arr[], int n)
    {
      
        int i = -1, j = 0;
        while (j != n) {
            if (arr[j] % 2 == 0)
            {
                i++;
      
                // Swapping even and
                // odd numbers
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
            j++;
        }
      
        // Printing segregated array
        for (int k = 0; k < n; k++)
             System.out.print(arr[k] + " ");
    }
    
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 1, 3, 2, 4, 7,
                            6, 9, 10 };
        int n = arr.length;
        arrayEvenAndOdd(arr, n);
    }
}
 
// This code is contributed by Sam007


Python3




# Python3 code to segregate even odd
# numbers in an array
 
# Function to segregate even odd numbers
def arrayEvenAndOdd(arr,n):
    i = -1
    j= 0
    while (j!=n):
        if (arr[j] % 2 ==0):
            i = i+1
 
            # Swapping even and odd numbers
            arr[i],arr[j] = arr[j],arr[i]
             
        j = j+1
     
    # Printing segregated array
    for i in arr:
        print (str(i) + " " ,end='')
         
# Driver Code
if __name__=='__main__':
    arr = [ 1 ,3, 2, 4, 7, 6, 9, 10]
    n = len(arr)
    arrayEvenAndOdd(arr,n)
 
# This code was contributed by
# Yatin Gupta


C#




// C# code to segregate even odd
// numbers in an array
using System;
 
class GFG {
     
    // Function to segregate even
    // odd numbers
    static void arrayEvenAndOdd(
                  int []arr, int n)
    {
     
        int i = -1, j = 0;
        while (j != n) {
            if (arr[j] % 2 == 0)
            {
                i++;
     
                // Swapping even and
                // odd numbers
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
            j++;
        }
     
        // Printing segregated array
        for (int k = 0; k < n; k++)
            Console.Write(arr[k] + " ");
    }
 
    // Driver code   
    static void Main()
    {
        int []arr = { 1, 3, 2, 4, 7,
                            6, 9, 10 };
        int n = arr.Length;
        arrayEvenAndOdd(arr, n);
    }
}
 
// This code is contributed by Sam007


PHP




<?php
// PHP code to segregate even odd
// numbers in an array
 
// Function to segregate
// even odd numbers
function arrayEvenAndOdd($arr, $n)
{
    $i = -1;
    $j = 0;
    $t;
    while ($j != $n)
    {
        if ($arr[$j] % 2 == 0)
        {
            $i++;
 
            // Swapping even and
            // odd numbers
            $x = $arr[$i];
            $arr[$i] = $arr[$j];
            $arr[$j] = $x;
        }
        $j++;
    }
 
    // Printing segregated
    // array
    for ($i = 0; $i < $n; $i++)
        echo $arr[$i] . " ";
}
 
    // Driver code
    $arr = array(1, 3, 2, 4, 7, 6, 9, 10);
    $n = sizeof($arr);
    arrayEvenAndOdd($arr, $n);
 
// This code is contributed by Anuj_67
?>


Javascript




<script>
// JavaScript code to segregate even odd
// numbers in an array
 
// Function to segregate even odd numbers
function arrayEvenAndOdd(arr, n)
{
 
    let i = -1, j = 0;
    let t;
    while (j != n) {
        if (arr[j] % 2 == 0) {
            i++;
 
            // Swapping even and odd numbers
                let temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
        }
        j++;
    }
 
    // Printing segregated array
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}
 
// Driver code
 
    let arr = [ 1, 3, 2, 4, 7, 6, 9, 10 ];
    let n = arr.length;
    arrayEvenAndOdd(arr, n);
 
// This code is contributed by Surbhi Tyagi
 
</script>


Output

2 4 6 10 7 1 9 3 

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




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