Segregate even and odd numbers | Set 3

14 Jun, 2022
Given an array arr[] of integers, segregate even and odd numbers in the array. Such that all the even numbers should be present first, and then the odd numbers.
Examples:  
Input: arr[] = 1 9 5 3 2 6 7 11Output: 2 6 5 3 1 9 7 11
Input: arr[] = 1 3 2 4 7 6 9 10Output:  2 4 6 10 7 1 9 3
We have discussed two different approaches in below posts: 
Segregate Even and Odd numbersSegregate even and odd numbers | Set 2
Segregate Even and Odd numbers
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 :
Create a temporary array A of size n and an integer index which will keep the index of elements inserted .Initialize index with zero and iterate over the original array and if even number is found then put that number at A[index] and then increment the value of index .Again iterate over array and if an odd number is found then put it in A[index] and then increment the value of index.Iterate over the temporary array A and copy its values in the original array.
Create a temporary array A of size n and an integer index which will keep the index of elements inserted .
Initialize index with zero and iterate over the original array and if even number is found then put that number at A[index] and then increment the value of index .
Again iterate over array and if an odd number is found then put it in A[index] and then increment the value of index.
Iterate over the temporary array A and copy its values in the original array.
C++
Java
Python3
C#
Javascript
// C++ Implementation of the above approach#include <iostream>using namespace std;void arrayEvenAndOdd(int arr[], int n){    int a[n], index = 0;    for (int i = 0; i < n; i++)    {        if (arr[i] % 2 == 0)        {            a[index] = arr[i];            index++;        }    }    for (int i = 0; i < n; i++)    {        if (arr[i] % 2 != 0)        {            a[index] = arr[i];            index++;        }    }    for (int i = 0; i < n; i++)    {        cout << a[i] << " ";    }    cout << endl;} // Driver codeint 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 Implementation of the above approachimport java.io.*;class GFG{public static void arrayEvenAndOdd(int arr[], int n){    int[] a;    a = new int[n];    int index = 0;    for (int i = 0; i < n; i++)    {        if (arr[i] % 2 == 0)        {            a[index] = arr[i];            index++;        }    }    for (int i = 0; i < n; i++)    {        if (arr[i] % 2 != 0)        {            a[index] = arr[i];            index++;        }    }    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 implementation of the above approachdef arrayEvenAndOdd(arr,  n):         index = 0;    a = [0 for i in range(n)]         for i in range(n):        if (arr[i] % 2 == 0):            a[index] = arr[i]            ind += 1     for i in range(n):        if (arr[i] % 2 != 0):            a[index] = arr[i]            ind += 1     for i in range(n):        print(a[i], end = " ")             print() # Driver codearr = [ 1, 3, 2, 4, 7, 6, 9, 10 ]n = len(arr) # Function callarrayEvenAndOdd(arr, n) # This code is contributed by rohitsingh07052
// C# implementation of the above approachusing System; class GFG{     static void arrayEvenAndOdd(int[] arr, int n){    int[] a = new int[n];    int index = 0;         for(int i = 0; i < n; i++)    {        if (arr[i] % 2 == 0)        {            a[index] = arr[i];            index++;        }    }    for(int i = 0; i < n; i++)    {        if (arr[i] % 2 != 0)        {            a[index] = arr[i];            index++;        }    }    for(int i = 0; i < n; i++)    {        Console.Write(a[i] + " ");    }} // Driver codestatic 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
<script> // JavaScript Implementation of the above approach function arrayEvenAndOdd(arr, n){    let a= [];    let index = 0;    for (let i = 0; i < n; i++)    {        if (arr[i] % 2 == 0)        {            a[index] = arr[i];            index++;        }    }    for (let i = 0; i < n; i++)    {        if (arr[i] % 2 != 0)        {            a[index] = arr[i];            index++;        }    }    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 callarrayEvenAndOdd(arr, n); </script>
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’s Partition Scheme 
Maintain a pointer to the position before first odd element in the array.Traverse the array and if even number is encountered then swap it with the first odd element.Continue the traversal.
Maintain a pointer to the position before first odd element in the array.
Traverse the array and if even number is encountered then swap it with the first odd element.
Continue the traversal.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
PHP
Javascript
// CPP code to segregate even odd// numbers in an array#include <bits/stdc++.h>using namespace std; // Function to segregate even odd numbersvoid 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 codeint main(){    int arr[] = { 1, 3, 2, 4, 7, 6, 9, 10 };    int n = sizeof(arr) / sizeof(int);    arrayEvenAndOdd(arr, n);    return 0;}
// java code to segregate even odd// numbers in an arraypublic 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 code to segregate even odd# numbers in an array # Function to segregate even odd numbersdef 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 Codeif __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# code to segregate even odd// numbers in an arrayusing 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 code to segregate even odd// numbers in an array // Function to segregate// even odd numbersfunction 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?>
<script>// JavaScript code to segregate even odd// numbers in an array // Function to segregate even odd numbersfunction 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>
2 4 6 10 7 1 9 3 
Time Complexity : O(n) Auxiliary Space : O(1) 
Sam007
vt_m
YatinGupta
nichil942singh
surbhityagi15
rohitsingh07052
lokeshpotta20
tarakki100
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