Even numbers at even index and odd numbers at odd index

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

// C++ implementation to arrange
// odd and even numbers
#include <bits/stdc++.h>
using namespace std;

// function to arrange odd and even numbers
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)
            swap (arr[evenInd], arr[oddInd]);

        else
            break;
    }
}

// function to print the array
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// Driver program to test above
int 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

// Java implementation to
// arrange odd and even numbers

import java.util.*;
import java.lang.*;

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++)
        System.out.print(arr[i] + " ");
}

// Driver function
public 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

# 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 array
def printArray(arr,  n):
    for i in range(0,n):
        print(arr[i] , "",end="")


# Driver function
def 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#

// C# implementation to
// arrange odd and even numbers
using 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

Javascript

<script>

// Javascript implementation to arrange
// odd and even numbers

// function to arrange odd and even numbers
function 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 array
function 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.

Approach: Two-Pointer Swap

The steps to solve the problem using this approach are as follows:

1. Initialize two pointers i and j to 0 and 1, respectively. These pointers will help us to keep track of the current indices of the elements in the array that we are processing.
2. Loop through the array until i or j reaches the end of the array.
3. If the element at index i is odd and the element at index j is even, swap them. This will ensure that the even number is moved to an even index and odd number is moved to an odd index.
4. Increment both i and j by 2 to move them to the next odd and even index respectively.
5. If the element at index i is already even, increment i by 2.
6. Similarly, if the element at index j is already odd, increment j by 2.
7. Once we have looped through the entire array, the even numbers will be at even indices and odd numbers will be at odd indices.
8. Return the modified array.

C++

#include <iostream>
#include <vector>

using namespace std;

vector<int> arrange_array(vector<int> arr) {
    int i = 0;
    int j = 1;
    int n = arr.size();
    while (i < n && j < n) {
        if (arr[i] % 2 != 0 && arr[j] % 2 == 0) {
            swap(arr[i], arr[j]);
            i += 2;
            j += 2;
        }
        else {
            if (arr[i] % 2 == 0) {
                i += 2;
            }
            if (arr[j] % 2 != 0) {
                j += 2;
            }
        }
    }
    return arr;
}

int main() {
    // Example 1
    vector<int> arr1 = {3, 6, 12, 1, 5, 8};
    vector<int> arranged_arr1 = arrange_array(arr1);
    for (auto i : arranged_arr1) {
        cout << i << " ";
    }
    cout << endl;

    // Example 2
    vector<int> arr2 = {10, 9, 7, 18, 13, 19, 4, 20, 21, 14};
    vector<int> arranged_arr2 = arrange_array(arr2);
    for (auto i : arranged_arr2) {
        cout << i << " ";
    }
    cout << endl;

    return 0;
}

Java

import java.util.Arrays;

public class Main {
    public static void main(String[] args)
    {
        int[] arr1 = { 3, 6, 12, 1, 5, 8 };
        int[] arrangedArr1 = arrangeArray(arr1);
        System.out.println(Arrays.toString(arrangedArr1));

        int[] arr2
            = { 10, 9, 7, 18, 13, 19, 4, 20, 21, 14 };
        int[] arrangedArr2 = arrangeArray(arr2);
        System.out.println(Arrays.toString(arrangedArr2));
    }

    // Function to arrange the array such that odd numbers
    // are at odd indices and even numbers are at even
    // indices
    public static int[] arrangeArray(int[] arr)
    {
        int i = 0; // starting index for odd numbers
        int j = 1; // starting index for even numbers
        int n = arr.length; // length of the array
        while (i < n && j < n) {
            // if the number at i is odd and the number at j
            // is even, swap them and move to next pair of
            // indices
            if (arr[i] % 2 != 0 && arr[j] % 2 == 0) {
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
                i += 2; // move to next odd index
                j += 2; // move to next even index
            }
            else {
                // if the number at i is even, move to next
                // odd index
                if (arr[i] % 2 == 0) {
                    i += 2;
                }
                // if the number at j is odd, move to next
                // even index
                if (arr[j] % 2 != 0) {
                    j += 2;
                }
            }
        }
        return arr; // return the arranged array
    }
}

Python3

def arrange_array(arr):
    i = 0
    j = 1
    n = len(arr)
    while i < n and j < n:
        if arr[i] % 2 != 0 and arr[j] % 2 == 0:
            arr[i], arr[j] = arr[j], arr[i]
            i += 2
            j += 2
        else:
            if arr[i] % 2 == 0:
                i += 2
            if arr[j] % 2 != 0:
                j += 2
    return arr

# Example 1
arr = [3, 6, 12, 1, 5, 8]
arranged_arr = arrange_array(arr)
print(arranged_arr)   # Output: [6, 3, 12, 1, 8, 5]

#Example
arr = [10, 9, 7, 18, 13, 19, 4, 20, 21, 14]
arranged_arr = arrange_array(arr)
print(arranged_arr)   # Output: [10, 9, 18, 7, 20, 19, 4, 13, 14, 21]

Javascript

function arrangeArray(arr) {
    let i = 0; // starting index for odd numbers
    let j = 1; // starting index for even numbers
    let n = arr.length; // length of the array
    while (i < n && j < n) {
        // if the number at i is odd and the number at j is even, swap them and move to next pair of indices
        if (arr[i] % 2 !== 0 && arr[j] % 2 === 0) {
            [arr[i], arr[j]] = [arr[j], arr[i]]; // swapping values
            i += 2; // move to next odd index
            j += 2; // move to next even index
        } else {
            // if the number at i is even, move to next odd index
            if (arr[i] % 2 === 0) {
                i += 2;
            }
            // if the number at j is odd, move to next even index
            if (arr[j] % 2 !== 0) {
                j += 2;
            }
        }
    }
    return arr; // return the arranged array
}

// Example 1
let arr1 = [3, 6, 12, 1, 5, 8];
let arrangedArr1 = arrangeArray(arr1);
console.log(arrangedArr1.join(" "));

// Example 2
let arr2 = [10, 9, 7, 18, 13, 19, 4, 20, 21, 14];
let arrangedArr2 = arrangeArray(arr2);
console.log(arrangedArr2.join(" "));

C#

using System;
using System.Collections.Generic;

class Program
{
    static List<int> arrangeArray(List<int> arr)
    {
        int i = 0;
        int j = 1;
        int n = arr.Count;
        while (i < n && j < n)
        {
            if (arr[i] % 2 != 0 && arr[j] % 2 == 0)
            {
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
                i += 2;
                j += 2;
            }
            else
            {
                if (arr[i] % 2 == 0)
                {
                    i += 2;
                }
                if (arr[j] % 2 != 0)
                {
                    j += 2;
                }
            }
        }
        return arr;
    }

    static void Main(string[] args)
    {
        // Example 1
        List<int> arr1 = new List<int> { 3, 6, 12, 1, 5, 8 };
        List<int> arranged_arr1 = arrangeArray(arr1);
        foreach (int i in arranged_arr1)
        {
            Console.Write(i + " ");
        }
        Console.WriteLine();

        // Example 2
        List<int> arr2 = new List<int> { 10, 9, 7, 18, 13, 19, 4, 20, 21, 14 };
        List<int> arranged_arr2 = arrangeArray(arr2);
        foreach (int i in arranged_arr2)
        {
            Console.Write(i + " ");
        }
        Console.WriteLine();

        Console.ReadLine();
    }
}

Output

[6, 3, 12, 1, 8, 5]
[10, 9, 18, 7, 20, 19, 4, 13, 14, 21]

This algorithm runs in O(n) time complexity and O(1) space complexity

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