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 partsbool 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 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; } 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 False# Driver codeif __name__ == '__main__': N = 8 # Function call ans = evenParts(N) if ans: print("YES") else: print("NO") |
C#
// C# Implementationusing 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 partsfunction 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 calllet ans = evenParts(N);if (ans) console.log("YES");else console.log("NO"); |
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 partsbool 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 codeint 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 approachimport 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 partsdef 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 codeN = 8# Function callans = evenParts(N)if (ans): print("YES")else: print("NO")# This code is contributed by Saurabh Jaiswal. |
C#
// C# code to implement above approachusing 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> |
YES
Time Complexity: O(1)
Auxiliary Space: O(1)
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