Divide a number into two unequal even parts

Namachila November 19, 2025

0 0 5 minutes read

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.

Brute Force Approach:

We iterate over all possible values of i from 1 to N-1, and check if i and (N-i) are both even and unequal. If we find such a pair of values, we return true, indicating that N can be divided into two unequal even parts. If we don’t find such a pair of values, we return false, indicating that N cannot be divided into two unequal even parts.

Implementation of the above approach:

C++

#include<iostream>
using namespace std;
 
// Function to check if N can be divided
// into two unequal even parts
bool evenParts(int N)
{
    for (int i = 1; i < N; i++) {
        if (i % 2 == 0 && (N-i) % 2 == 0 && i != (N-i)) {
            return true;
        }
    }
    return false;
}
 
// Driver code
int main(){
    int N = 8;
     
    // Function call
    bool ans = evenParts(N);
     
    if(ans)
        cout << "YES" << '\n';
    else
        cout << "NO" << '\n';
     
    return 0;
}

Java

public class EvenParts {
// Function to check if N can be divided
// into two unequal even parts
    public static boolean evenParts(int N) {
        for (int i = 1; i < N; i++) {
            if (i % 2 == 0 && (N - i) % 2 == 0 && i != (N - i)) {
                return true;
            }
        }
        return false;
    }
 
    public static void main(String[] args) {
        int N = 8; // Sample input
        boolean ans = evenParts(N);
        if (ans) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}

Python3

# Function to check if N can be divided
# into two unequal even parts
def evenParts(N):
    for i in range(1, N):
        if i % 2 == 0 and (N - i) % 2 == 0 and i != (N - i):
            return True
    return False
 
# Driver code
if __name__ == '__main__':
    N = 8
 
    # Function call
    ans = evenParts(N)
 
    if ans:
        print("YES")
    else:
        print("NO")

C#

// C# Implementation
using System;
 
class GFG {
    // Function to check if N can be divided 
    // into two unequal even parts
    static bool evenParts(int N) {
        for (int i = 1; i < N; i++) {
            if (i % 2 == 0 && (N - i) % 2 == 0 && i != (N - i)) {
                return true;
            }
        }
        return false;
    }
 
    // Driver code
    static void Main(string[] args) {
        int N = 8;
 
        // Function call
        bool ans = evenParts(N);
 
        if (ans)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}
 
// This code is contributed by Utkarsh Kumar

Javascript

// Function to check if N can be divided into two unequal even parts
function evenParts(N) {
    for (let i = 1; i < N; i++) {
        if (i % 2 == 0 && (N - i) % 2 == 0 && i != (N - i)) {
            return true;
        }
    }
    return false;
}
 
let N = 8;
 
// Function call
let ans = evenParts(N);
 
if (ans)
    console.log("YES");
else
    console.log("NO");

Output

YES

Time Complexity: O(n)

Space Complexity: O(1)

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:

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

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