Even numbers at even index and odd numbers at odd index

31 May, 2022
Given an array of size n containing equal number of odd and even numbers. The problem is to arrange the numbers in such a way that all the even numbers get the even index and odd numbers get the odd index. Required auxiliary space is O(1).Examples :  
Input : arr[] = {3, 6, 12, 1, 5, 8}
Output : 6 3 12 1 8 5 

Input : arr[] = {10, 9, 7, 18, 13, 19, 4, 20, 21, 14}
Output : 10 9 18 7 20 19 4 13 14 21 
Source: Amazon Interview Experience | Set 410. 
Approach :  
Start from the left and keep two index one for even position and other for odd positions.
Traverse these index from left.
At even position there should be even number and at odd positions, there should be odd number.
Whenever there is mismatch , we swap the values at odd and even index.
Below is the implementation of the above approach : 
CPP
Java
Python3
C#
Javascript
// C++ implementation to arrange// odd and even numbers#include <bits/stdc++.h>using namespace std; // function to arrange odd and even numbersvoid arrangeOddAndEven(int arr[], int n){   int oddInd = 1;    int evenInd = 0;    while (true)    {        while (evenInd < n && arr[evenInd] % 2 == 0)            evenInd += 2;                     while (oddInd < n && arr[oddInd] % 2 == 1)            oddInd += 2;                     if (evenInd < n && oddInd < n)            swap (arr[evenInd], arr[oddInd]);                     else            break;    }} // function to print the arrayvoid printArray(int arr[], int n){    for (int i = 0; i < n; i++)        cout << arr[i] << " ";} // Driver program to test aboveint main(){    int arr[] = { 3, 6, 12, 1, 5, 8 };    int n = sizeof(arr) / sizeof(arr[0]);     cout << "Original Array: ";    printArray(arr, n);     arrangeOddAndEven(arr, n);     cout << "\nModified Array: ";    printArray(arr, n);     return 0;}
// Java implementation to// arrange odd and even numbers import java.util.*;import java.lang.*; class GfG { // function to arrange// odd and even numberspublic static void arrangeOddAndEven(int arr[], int n){    int oddInd = 1;    int evenInd = 0;    while (true)    {        while (evenInd < n && arr[evenInd] % 2 == 0)            evenInd += 2;                     while (oddInd < n && arr[oddInd] % 2 == 1)            oddInd += 2;                     if (evenInd < n && oddInd < n)            {                int temp = arr[evenInd];                arr[evenInd] = arr[oddInd];                arr[oddInd] = temp;            }                     else            break;    }} // function to print the arraypublic static void printArray(int arr[], int n){    for (int i = 0; i < n; i++)        System.out.print(arr[i] + " ");}     // Driver functionpublic static void main(String argc[]){    int arr[] = { 3, 6, 12, 1, 5, 8 };    int n = 6;     System.out.print("Original Array: ");    printArray(arr, n);     arrangeOddAndEven(arr, n);     System.out.print("\nModified Array: ");    printArray(arr, n);}} // This code is contributed by Sagar Shukla
     # Python3 implementation to# arrange odd and even numbers def arrangeOddAndEven(arr,  n):         oddInd = 1    evenInd = 0    while (True):                 while (evenInd < n and arr[evenInd] % 2 == 0):            evenInd += 2                      while (oddInd < n and arr[oddInd] % 2 == 1):            oddInd += 2                      if (evenInd < n and oddInd < n):                temp = arr[evenInd]                arr[evenInd] = arr[oddInd]                arr[oddInd] = temp;                      else:            break  # function to print the arraydef printArray(arr,  n):    for i in range(0,n):        print(arr[i] , "",end="")       # Driver functiondef main():    arr = [ 3, 6, 12, 1, 5, 8 ]    n = 6      print("Original Array: ",end="")    printArray(arr, n)      arrangeOddAndEven(arr, n)      print("\nModified Array: ",end="")    printArray(arr, n)     if __name__ == '__main__':    main()# This code is contributed by 29AjayKumar
// C# implementation to// arrange odd and even numbersusing System; class GFG {     // function to arrange    // odd and even numbers    public static void arrangeOddAndEven(int[] arr, int n)    {     int oddInd = 1;     int evenInd = 0;    while (true)    {        while (evenInd < n && arr[evenInd] % 2 == 0)            evenInd += 2;                      while (oddInd < n && arr[oddInd] % 2 == 1)            oddInd += 2;                      if (evenInd < n && oddInd < n)            {                int temp = arr[evenInd];                arr[evenInd] = arr[oddInd];                arr[oddInd] = temp;            }                      else            break;    }    }     // function to print the array    public static void printArray(int[] arr, int n)    {        for (int i = 0; i < n; i++)            Console.Write(arr[i] + " ");    }     // Driver function    public static void Main()    {        int[] arr = { 3, 6, 12, 1, 5, 8 };        int n = 6;         Console.Write("Original Array: ");        printArray(arr, n);         arrangeOddAndEven(arr, n);         Console.Write("\nModified Array: ");        printArray(arr, n);    }} // This code is contributed by Sam007
<script> // Javascript implementation to arrange// odd and even numbers // function to arrange odd and even numbersfunction arrangeOddAndEven(arr, n){    let oddInd = 1;    let evenInd = 0;    while (true)    {        while (evenInd < n && arr[evenInd] % 2 == 0)            evenInd += 2;                     while (oddInd < n && arr[oddInd] % 2 == 1)            oddInd += 2;                     if (evenInd < n && oddInd < n)        {            let temp;            temp = arr[evenInd];            arr[evenInd] = arr[oddInd];            arr[oddInd] = temp;        }           else            break;    }} // function to print the arrayfunction printArray(arr, n){    for (let i = 0; i < n; i++)        document.write(arr[i] + " ");} // Driver program to test above    let arr = [ 3, 6, 12, 1, 5, 8 ];    let n = arr.length;     document.write("Original Array: ");    printArray(arr, n);     arrangeOddAndEven(arr, n);     document.write("<br>" + "Modified Array: ");    printArray(arr, n); // This code is contributed by Mayank Tyagi </script>
Output : 
Original Array: 3 6 12 1 5 8 
Modified Array: 6 3 12 1 8 5 
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
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