Divide a number into two unequal even parts

Given a positive integer N. The task is to decide whether the integer can be divided into two unequal positive even parts or not.

Examples:

Input: N = 8
Output: YES
Explanation: 8 can be divided into two different even parts i.e. 2 and 6.

Input: N = 5
Output: NO
Explanation: 5 can not be divided into two even parts in any way.

Input:  N = 4
Output: NO
Explanation: 4 can be divided into two even parts, 2 and 2. Since the numbers are equal, the output is NO.

Prerequisites: Knowledge of if-else conditional statements.

Approach: The core concept of the problem lies in the following observation:

The sum of any two even numbers is always even. Conversely any even number can be expressed as sum of two even numbers.

But here is two exceptions

• The number 2 is an exception here. It can only be expressed as the sum of two odd numbers (1 + 1).
• The number 4 can only be expressed as the sum of equal even numbers (2 + 2).

Hence, it is possible to express N as the sum of two even numbers only if N is even and not equal to 2 or 4. If N is odd, it is impossible to divide it into two even parts. Follow the steps mentioned below:

1. Check if N = 2 or N = 4.
2. If yes, then print NO.
3. Else check if N is even (i.e. a multiple of 2)
4. If yes, then print YES.
5. Else, print NO.

Below is the implementation of the above approach.

C++

// C++ code to implement above approach
#include<iostream>
using namespace std;

// Function to check if N can be divided
// into two unequal even parts
bool evenParts(int N)
{
    // Check if N is equal to 2 or 4
    if(N == 2 || N == 4)
        return false;

    // Check if N is even
    if(N % 2 == 0)
        return true;
    else
        return false;
}

//Driver code
int main(){
   int N = 8;

   // Function call
   bool ans = evenParts(N);

   if(ans)
       std::cout << "YES" << '\n';
   else
       std::cout << "NO" << '\n';

   return 0;
}

Java

// Java code to implement above approach
import java.util.*;
public class GFG {

  // Function to check if N can be divided
  // into two unequal even parts
  static boolean evenParts(int N)
  {

    // Check if N is equal to 2 or 4
    if(N == 2 || N == 4)
      return false;

    // Check if N is even
    if(N % 2 == 0)
      return true;
    else
      return false;
  }

  // Driver code
  public static void main(String args[])
  {
    int N = 8;

    // Function call
    boolean ans = evenParts(N);

    if(ans)
      System.out.println("YES");
    else
      System.out.println("NO");

  }
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach

# Function to check if N can be divided
# into two unequal even parts
def evenParts(N):

    # Check if N is equal to 2 or 4
    if (N == 2 or N == 4):
        return False

    # Check if N is even
    if (N % 2 == 0):
        return True
    else:
        return False

# Driver code
N = 8

# Function call
ans = evenParts(N)
if (ans):
    print("YES")
else:
    print("NO")

# This code is contributed by Saurabh Jaiswal.

C#

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

  // Function to check if N can be divided
  // into two unequal even parts
  static bool evenParts(int N)
  {

    // Check if N is equal to 2 or 4
    if(N == 2 || N == 4)
      return false;

    // Check if N is even
    if(N % 2 == 0)
      return true;
    else
      return false;
  }

  // Driver code
  public static void Main()
  {
    int N = 8;

    // Function call
    bool ans = evenParts(N);

    if(ans)
      Console.Write("YES" + '\n');
    else
      Console.Write("NO" + '\n');

  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
       // JavaScript code for the above approach

       // Function to check if N can be divided
       // into two unequal even parts
       function evenParts(N)
       {

           // Check if N is equal to 2 or 4
           if (N == 2 || N == 4)
               return false;

           // Check if N is even
           if (N % 2 == 0)
               return true;
           else
               return false;
       }

       // Driver code
       let N = 8;

       // Function call
       let ans = evenParts(N);
       if (ans)
           document.write("YES" + '<br>')
       else
           document.write("NO" + '<br>')

 // This code is contributed by Potta Lokesh
   </script>

Output

YES

Time Complexity: O(1)
Auxiliary Space: O(1)