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: Â 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.
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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;Â
// Function to check if N can be divided// into two unequal even partsbool evenParts(int N){Â Â Â Â for (int i = 1; i < N; i++) {Â Â Â Â Â Â Â Â if (i % 2 == 0 && (N-i) % 2 == 0 && i != (N-i)) {Â Â Â Â Â Â Â Â Â Â Â Â return true;Â Â Â Â Â Â Â Â }Â Â Â Â }Â Â Â Â return false;}Â
// Driver codeint 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;    }Â
    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 partsdef evenParts(N):    for i in range(1, N):        if i % 2 == 0 and (N - i) % 2 == 0 and i != (N - i):            return True    return FalseÂ
# Driver codeif __name__ == '__main__':Â Â Â Â N = 8Â
    # Function call    ans = evenParts(N)Â
    if ans:        print("YES")    else:        print("NO") |
C#
// C# Implementationusing System;Â
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;    }Â
    // Driver code    static void Main(string[] args) {        int N = 8;Â
        // Function call        bool ans = evenParts(N);Â
        if (ans)            Console.WriteLine("YES");        else            Console.WriteLine("NO");    }}Â
// This code is contributed by Utkarsh Kumar |
Javascript
// Function to check if N can be divided into two unequal even partsfunction evenParts(N) {Â Â Â Â for (let i = 1; i < N; i++) {Â Â Â Â Â Â Â Â if (i % 2 == 0 && (N - i) % 2 == 0 && i != (N - i)) {Â Â Â Â Â Â Â Â Â Â Â Â return true;Â Â Â Â Â Â Â Â }Â Â Â Â }Â Â Â Â return false;}Â
let N = 8;Â
// Function calllet ans = evenParts(N);Â
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;Â
// Function to check if N can be divided // into two unequal even partsbool evenParts(int N){Â Â Â Â Â Â // Check if N is equal to 2 or 4Â Â Â Â Â if(N == 2 || N == 4)Â Â Â Â Â Â Â Â return false;Â
    // Check if N is even    if(N % 2 == 0)        return true;    else        return false;}  //Driver codeint main(){   int N = 8;        // Function call   bool ans = evenParts(N);Â
   if(ans)       std::cout << "YES" << '\n';   else       std::cout << "NO" << '\n';     return 0;} |
Java
// Java code to implement above approachimport java.util.*;public class GFG {Â
  // Function to check if N can be divided   // into two unequal even parts  static boolean evenParts(int N)  {  Â
    // Check if N is equal to 2 or 4     if(N == 2 || N == 4)      return false;Â
    // Check if N is even    if(N % 2 == 0)      return true;    else      return false;  }Â
  // Driver code  public static void main(String args[])  {    int N = 8;Â
    // Function call    boolean ans = evenParts(N);Â
    if(ans)      System.out.println("YES");    else      System.out.println("NO");Â
  }}Â
// This code is contributed by Samim Hossain Mondal. |
Python3
# Python code for the above approachÂ
# Function to check if N can be divided# into two unequal even partsdef evenParts(N):Â
    # Check if N is equal to 2 or 4    if (N == 2 or N == 4):        return FalseÂ
    # Check if N is even    if (N % 2 == 0):        return True    else:        return FalseÂ
# Driver codeN = 8Â
# Function callans = evenParts(N)if (ans):Â Â Â Â print("YES")else:Â Â Â Â print("NO")Â
# This code is contributed by Saurabh Jaiswal. |
C#
// C# code to implement above approachusing System;class GFG {Â
  // 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;Â
    // Check if N is even    if(N % 2 == 0)      return true;    else      return false;  }Â
  // Driver code  public static void Main()  {    int N = 8;Â
    // Function call    bool ans = evenParts(N);Â
    if(ans)      Console.Write("YES" + '\n');    else      Console.Write("NO" + '\n');Â
  }}Â
// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>Â Â Â Â Â Â Â // JavaScript code for the above approachÂ
       // 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;Â
           // Check if N is even           if (N % 2 == 0)               return true;           else               return false;       }Â
       // Driver code       let N = 8;Â
       // Function call       let ans = evenParts(N);       if (ans)           document.write("YES" + '<br>')       else           document.write("NO" + '<br>')Â
 // This code is contributed by Potta Lokesh   </script> |
Â
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Output
YES
Â
Time Complexity: O(1)
Auxiliary Space: O(1)
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