Given a number N, the task is to check if this number can be divided into 2 even parts.
Examples:
Input: N = 8
Output: YES
Explanation: 8 can be divided into two even parts in two ways, 2, 6 or 4, 4 as both are even.Input: N = 5
Output: NO
Naive approach: Check all number pairs upto N, such that they both are even and they both sum to N. If possible, print Yes, else No.
Code-
C++
// C++ code to implement above approach#include <bits/stdc++.h>using namespace std;// Divide number into 2 even partsbool divNum(int n){ for(int i=1;i<n;i++){ for(int j=1;j<n;j++){ if((i%2==0) && (j%2==0) && (i+j==n)){ return true; } } } return false;}// Driven Programint main(){ int n = 8; cout << (divNum(n) ? "YES" : "NO") << endl; return 0;} |
Java
import java.util.*;class Main {// Divide number into 2 even partsstatic boolean divNum(int n) { for (int i = 1; i < n; i++) { for (int j = 1; j < n; j++) { if ((i % 2 == 0) && (j % 2 == 0) && (i + j == n)) { return true; } } } return false;}// Driven Programpublic static void main(String[] args) { int n = 8; System.out.println(divNum(n) ? "YES" : "NO");}} |
Python3
# Python code to implement above approach# Divide number into 2 even partsdef divNum(n): for i in range(1, n): for j in range(1, n): if i % 2 == 0 and j % 2 == 0 and i + j == n: return True return False# Driven Programif __name__ == '__main__': n = 8 print("YES" if divNum(n) else "NO") |
C#
using System;class Program{ // Divide number into 2 even parts static bool DivNum(int n) { for (int i = 1; i < n; i++) { for (int j = 1; j < n; j++) { if ((i % 2 == 0) && (j % 2 == 0) && (i + j == n)) { return true; } } } return false; } // Driven Program static void Main(string[] args) { int n = 8; Console.WriteLine(DivNum(n) ? "YES" : "NO"); }} |
Javascript
// Javascript code to implement above approach// Divide number into 2 even partsfunction divNum(n) { for (let i = 1; i < n; i++) {for (let j = 1; j < n; j++) { if (i % 2 === 0 && j % 2 === 0 && i + j === n) { return true; }} } return false;}// Driven Program const n = 8; console.log(divNum(n) ? "YES" : "NO"); // This code is contributed by Utkarsh Kumar |
Output
YES
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient approach: Upon observation, it can be noticed that any even number can be expressed as sum of two even numbers except for 2 and no two even number can sum up to form an odd number.
Below is the implementation of the above approach.
C++
// C++ code to implement above approach#include <bits/stdc++.h>using namespace std;// Divide number into 2 even partsbool divNum(int n){ return (n <= 2 ? false : (n % 2 == 0 ? true : false));}// Driven Programint main(){ int n = 8; cout << (divNum(n) ? "YES" : "NO") << endl; return 0;} |
Java
// Java code to implement above approachimport java.util.*;public class GFG { // Divide number into 2 even parts static boolean divNum(int n) { return (n <= 2 ? false : (n % 2 == 0 ? true : false)); } // Driven Program public static void main(String args[]) { int n = 8; System.out.println(divNum(n) ? "YES" : "NO"); }}// This code is contributed by Samim Hossain Mondal. |
Python
# Python program for above approach# Divide number into 2 even partsdef divNum(n): ans = False if n <= 2 else True if n % 2 == 0 else False return ans# Driver Coden = 8if(divNum(n) == True): print("YES")else: print("NO")# This code is contributed by Samim Hossain Mondal. |
C#
// C# code to implement above approachusing System;class GFG { // Divide number into 2 even parts static bool divNum(int n) { return (n <= 2 ? false : (n % 2 == 0 ? true : false)); } // Driven Program public static void Main() { int n = 8; Console.WriteLine(divNum(n) ? "YES" : "NO"); }}// This code is contributed by ukasp. |
Javascript
<script>// Javascript code to implement above approach// Divide number into 2 even partsfunction divNum(n){ return (n <= 2 ? false : (n % 2 == 0 ? true : false));}// Driven Programlet n = 8;document.write((divNum(n) ? "YES" : "NO") + "\n");// This code is contributed by Samim Hossin Mondal.</script> |
Output
YES
Time Complexity: O(1)
Auxiliary Space: O(1)
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