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

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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.

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