Segregate even and odd numbers | Set 2

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:

Input : 1 9 5 3 2 6 7 11
Output : 6 2 3 5 2 9 11 1

Input : 1 3 2 4 7 6 9 10
Output : 10 2 6 4 7 9 3 1

We have discussed one approach in Segregate Even and Odd numbers. In this post, a different simpler approach is discussed that uses an extra array.

Approach :
1. Start two pointers from left and right of the array.
2. Create a new array of same size as given.
3. If the element at left or right is even then put it in front of the array else at the end.

C++

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// 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 b[n];  // To store result
  int k = 0, l = n - 1, i, j;
  for (i = 0, j = n - 1; i < j; i++, j--) {
  
    if (arr[i] % 2 == 0) {
      b[k] = arr[i];
      k++;
    } else {
      b[l] = arr[i];
      l--;
    }
  
    if (arr[j] % 2 == 0) {
      b[k] = arr[j];
      k++;
    } else {
      b[l] = arr[j];
      l--;
    }
  }
  
  // for i == j in case of odd length
  b[i] = arr[i];
  
  // Printing segregated array
  for (int i = 0; i < n; i++) 
     cout << b[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;
}

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Java

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// Java code to segregate even odd
// numbers in an array
import java.util.Arrays;
  
class arr
{
    // Function to segregate even odd numbers
    static void arrayEvenAndOdd(int arr[], int n) 
    {
        // To store result
        int b[] = new int[n]; 
        int k = 0, l = n - 1, i, j;
        for (i = 0, j = n - 1; i < j; i++, j--) 
        {
          
            if (arr[i] % 2 == 0
            {
                b[k] = arr[i];
                k++;
            
            else 
            {
                b[l] = arr[i];
                l--;
            }
          
            if (arr[j] % 2 == 0)
            {
                b[k] = arr[j];
                k++;
            
            else 
            {
                b[l] = arr[j];
                l--;
            }
        }
      
        // for i == j in case of odd length
        b[i] = arr[i];
          
        // Printing segregated array
        for (i = 0; i < n; i++) 
        {
            System.out.print(b[i] + " ");
        }
    }
      
    // 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 Smitha Dinesh Semwal

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Python3

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# Python3 code to segregate even odd
# numbers in an array
  
# Function to segregate even odd numbers
def arrayEvenAndOdd(arr,n):  
  
    b = [0] * # To store result
    k = 0
    l = n - 1
    i = 0
    j = n-1
    while(i < j):
        if (arr[i] % 2 == 0):  
            b[k] = arr[i] 
            k+=1
        else:  
            b[l] = arr[i] 
            l-=1
        if (arr[j] % 2 == 0):  
            b[k] = arr[j] 
            k+=1
        else:  
            b[l] = arr[j] 
            l-=1 
        i+=1
        j-=1
       
   
  
    # for i == j in case of odd length
    b[i] = arr[i] 
  
    # Printing segregated array
    for i in range(0, n): 
        print(b[i],end=" ")  
   
  
# Driver code
arr =  [1, 3, 2, 4, 7, 6, 9, 10]  
n = len(arr)
  
arrayEvenAndOdd(arr, n)
  
# This code is contributed by
# Smitha Dinesh Semwal

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C#

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// 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) 
    {
          
        // To store result
        int []b = new int[n]; 
        int k = 0, l = n - 1, i, j;
        for (i = 0, j = n - 1; i < j; i++, j--) 
        {
          
            if (arr[i] % 2 == 0) 
            {
                b[k] = arr[i];
                k++;
            
            else
            {
                b[l] = arr[i];
                l--;
            }
          
            if (arr[j] % 2 == 0)
            {
                b[k] = arr[j];
                k++;
            
            else
            {
                b[l] = arr[j];
                l--;
            }
        }
      
        // for i == j in case of odd length
        b[i] = arr[i];
          
        // Printing segregated array
        for (i = 0; i < n; i++) 
        {
            Console.Write(b[i] + " ");
        }
    }
      
    // Driver code
    public 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 vt_m.

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Output:

10 2 6 4 7 9 3 1

Time Complexity : O(n/2)
Auxiliary Space : O(n)



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