Rearrange array such that all even-indexed elements in the Array is even

Given an array arr[], the task is to check if it is possible to rearrange the array in such a way that every even index(1-based indexing) contains an even number. If such a rearrangement is not possible, print “No”. Otherwise, print “Yes” and print a possible arrangement

Examples:

Input: arr[] = {2, 4, 8, 3, 1} 
Output:
Yes 
3 4 8 2 1
Input: arr[] = {3, 3, 11, 8} 
Output: No 
Explanation: Since, the array contains only one even element, all even indices cannot be filled with a even elements.

Approach:
Follow the steps below to solve the problem: 

• Count the total number of even elements present in the given array. If the count exceeds the total number of even indices in the given array, print “No”.
• Otherwise, print “Yes”. Now, traverse the array using two pointers, i and j, pointing to even and odd indices respectively.
• For any i^th index containing an odd element, iterate odd indices using j until an even element is encountered.
• Swap a[i] and a[j]
• Repeat the above steps until all even indices are filled with an even element.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if it the
// array can be rearranged such
// such that every even indices
// contains an even element
void checkPossible(int a[], int n)
{
    // Stores the count of even elements
    int even_no_count = 0;

    // Traverse array to count even numbers
    for (int i = 0; i < n; i++)
    {
        if (a[i] % 2 == 0)
            even_no_count++;
    }

    // If even_no_count exceeds
    // count of even indices
    if (n / 2 > even_no_count)
    {
        cout << "No" << endl;
        return;
    }

    cout << "Yes" << endl;
    int j = 0;
    for (int i = 1; i < n; i += 2)
    {
        if (a[i] % 2 == 0)
            continue;
        else
        {
            while (j < n && a[j] % 2 != 0)
                j += 2;

            a[i] += a[j];
            a[j] = a[i] - a[j];
            a[i] -= a[j];
        }
    }

    for (int i = 0; i < n; i++)
    {
        cout << a[i] << " ";
    }
}

// Driver Code
int main()
{

    int arr[] = { 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    checkPossible(arr, n);
    return 0;
}

// This code is contributed by gauravrajput1

Java

// Java Program to implement
// the above approach
import java.io.*;
import java.util.*;

class GFG {

    // Function to check if it the
    // array can be rearranged such
    // such that every even indices
    // contains an even element
    static void checkPossible(int a[])
    {
        // Stores the count of even elements
        int even_no_count = 0;

        // Traverse array to count even numbers
        for (int i = 0; i < a.length; i++) {

            if (a[i] % 2 == 0)
                even_no_count++;
        }

        // If even_no_count exceeds
        // count of even indices
        if (a.length / 2 > even_no_count) {
            System.out.println("No");
            return;
        }

        System.out.println("Yes");
        int j = 0;
        for (int i = 1; i < a.length; i += 2) {
            if (a[i] % 2 == 0)
                continue;
            else {
                while (j < a.length && a[j] % 2 != 0)
                    j += 2;

                a[i] += a[j];
                a[j] = a[i] - a[j];
                a[i] -= a[j];
            }
        }

        for (int i = 0; i < a.length; i++) {
            System.out.print(a[i] + " ");
        }
    }

    // Driver Code
    public static void main(String args[])
    {

        int arr[] = { 2, 3, 4, 5, 6, 7 };

        checkPossible(arr);
    }
}

Python3

# Python3 program to implement
# the above approach

# Function to check if it the
# array can be rearranged such
# such that every even indices
# contains an even element
def checkPossible(a, n):

    # Stores the count of even elements
    even_no_count = 0

    # Traverse array to count even numbers
    for i in range(n):
        if (a[i] % 2 == 0):
            even_no_count += 1

    # If even_no_count exceeds
    # count of even indices
    if (n // 2 > even_no_count):
        print("No")
        return

    print("Yes")
    j = 0

    for i in range(1, n, 2):
        if (a[i] % 2 == 0):
            continue

        else:
            while (j < n and a[j] % 2 != 0):
                j += 2

            a[i] += a[j]
            a[j] = a[i] - a[j]
            a[i] -= a[j]

    for i in range(n):
        print(a[i], end = " ")

# Driver Code
arr = [ 2, 3, 4, 5, 6, 7 ]
n = len(arr)

checkPossible(arr, n)

# This code is contributed by code_hunt

C#

// C# Program to implement
// the above approach
using System;
class GFG{

  // Function to check if it the
  // array can be rearranged such
  // such that every even indices
  // contains an even element
  static void checkPossible(int []a)
  {
    // Stores the count of even elements
    int even_no_count = 0;

    // Traverse array to count even numbers
    for (int i = 0; i < a.Length; i++)
    {
      if (a[i] % 2 == 0)
        even_no_count++;
    }

    // If even_no_count exceeds
    // count of even indices
    if (a.Length / 2 > even_no_count)
    {
      Console.WriteLine("No");
      return;
    }

    Console.WriteLine("Yes");
    int j = 0;
    for (int i = 1; i < a.Length; i += 2)
    {
      if (a[i] % 2 == 0)
        continue;
      else
      {
        while (j < a.Length && a[j] % 2 != 0)
          j += 2;

        a[i] += a[j];
        a[j] = a[i] - a[j];
        a[i] -= a[j];
      }
    }

    for (int i = 0; i < a.Length; i++)
    {
      Console.Write(a[i] + " ");
    }
  }

  // Driver Code
  public static void Main(String []args)
  {
    int []arr = { 2, 3, 4, 5, 6, 7 };

    checkPossible(arr);
  }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Java Script Program to implement
// the above approach


    // Function to check if it the
    // array can be rearranged such
    // such that every even indices
    // contains an even element
    function checkPossible(a)
    {
        // Stores the count of even elements
        let even_no_count = 0;

        // Traverse array to count even numbers
        for (let i = 0; i < a.length; i++) {

            if (a[i] % 2 == 0)
                even_no_count++;
        }

        // If even_no_count exceeds
        // count of even indices
        if (a.length / 2 > even_no_count) {
            document.write("No<br>");
            return;
        }

        document.write("Yes<br>");
        let j = 0;
        for (let i = 1; i < a.length; i += 2) {
            if (a[i] % 2 == 0)
                continue;
            else {
                while (j < a.length && a[j] % 2 != 0)
                    j += 2;

                a[i] += a[j];
                a[j] = a[i] - a[j];
                a[i] -= a[j];
            }
        }

        for (let i = 0; i < a.length; i++) {
            document.write(a[i] + " ");
        }
    }

    // Driver Code


        let arr = [ 2, 3, 4, 5, 6, 7 ];

        checkPossible(arr);

// This code is contributed by sravan kumar Gottumukkala
</script>

Output:

Yes
3 2 5 4 7 6

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.