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# Check if K distinct positive even integers with given sum exists or not

Given two integers N and K, the task is to find if there exist K distinct positive even integers such that their sum is equal to the given number N.

Examples:

Input: N = 16, K = 3
Output: Yes
Explanation: Three distinct positive even integers with sum 16 are 8, 6 and 2.
Since, three such numbers exist, print “YES”.

Input: N = 18 K = 4
Output: No
Explanation: Sum of four distinct positive even integers cannot be equal to 18. Hence, print “NO”.

Approach: The idea to solve this problem is to observe that if N is odd, then it will be impossible to get required N by K even numbers. If N is even, then find the sum of the first K even numbers and if their sum is less than or equal to N, then print “YES”. Otherwise, there does not exist K distinct even integers with sum equal to N.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std; // Function to find if the sum// of K even integers equals Nvoid isPossibleN(int N, int K){     // If N is odd, then its impossible    // to obtain N from K even numbers    if (N % 2 == 1) {        cout << "NO" << endl;        return;    }     // Sum first k even numbers    int sum = K * (K + 1);     // If sum is less than equal to N    if (sum <= N) {        cout << "YES" << endl;    }     // Otherwise    else {        cout << "No" << endl;    }     return;} // Driver Codeint main(){     int N = 16;    int K = 3;     isPossibleN(N, K);     return 0;}

## Java

 // Java program for the above approachimport java.util.*;class GFG{  static void isPossibleN(int N, int K){     // If N is odd, then its impossible    // to obtain N from K even numbers    if (N % 2 == 1) {        System.out.println("NO");        return;    }     // Sum first k even numbers    int sum = K * (K + 1);     // If sum is less than equal to N    if (sum <= N) {        System.out.println("YES");    }     // Otherwise    else {        System.out.println("No");    }} // Driver Codepublic static void main(String args[]){    int N = 16;    int K = 3;    isPossibleN(N, K);}} // This code is contributed by jana_sayantan.

## Python3

 # Python3 program for the above approach # Function to find if the sum# of K even integers equals Ndef isPossibleN(N, K) :      # If N is odd, then its impossible    # to obtain N from K even numbers    if (N % 2 == 1) :        print("NO")        return           # Sum first k even numbers    sum = K * (K + 1)      # If sum is less than equal to N    if (sum <= N) :        print("YES")           # Otherwise    else :        print("NO")      return   # Driver Code N = 16K = 3  isPossibleN(N, K)

## C#

 // C# implementation of the// above approachusing System;class GFG{   static void isPossibleN(int N, int K)  {     // If N is odd, then its impossible    // to obtain N from K even numbers    if (N % 2 == 1) {      Console.WriteLine("NO");      return;    }     // Sum first k even numbers    int sum = K * (K + 1);     // If sum is less than equal to N    if (sum <= N) {      Console.WriteLine("YES");    }     // Otherwise    else {      Console.WriteLine("No");    }  }   // Driver Code  public static void Main()  {    int N = 16;    int K = 3;    isPossibleN(N, K);  }} // This code is contributed by jana_sayantan.

## Javascript



Output

YES

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

Approach 2:

Here’s another approach to solve the problem:

• Take two integers, N and K.
• If K is greater than N/2, return “NO”, since we cannot obtain K even numbers whose sum is equal to N.
• If N is even and K is odd, return “NO”, since the sum of K even numbers will always be even.
• Otherwise, return “YES”.
Here are the codes for the same:

## C++

 #include using namespace std; string isPossibleN(int N, int K){    // Check if K is greater than N/2    if(K > N/2) return "NO";         // Check if N is even and K is odd    if(N % 2 == 0 && K % 2 == 1) return "YES";         return "NO";} int main(){    int N = 16;    int K = 3;    cout << isPossibleN(N, K);    return 0;}

## Java

 import java.util.*; public class Main {public static String isPossibleN(int N, int K) {// Check if K is greater than N/2if (K > N / 2)return "NO";      // Check if N is even and K is odd    if (N % 2 == 0 && K % 2 == 1)        return "YES";     return "NO";} public static void main(String[] args) {    int N = 16;    int K = 3;    System.out.println(isPossibleN(N, K));}}

## Python3

 def is_possible_N(N, K):    # Check if K is greater than N/2    if K > N//2:        return "NO"         # Check if N is even and K is odd    if N % 2 == 0 and K % 2 == 1:        return "YES"         return "NO" if __name__ == '__main__':    N = 16    K = 3    print(is_possible_N(N, K))

## C#

 using System; class GFG{    static string IsPossibleN(int N, int K)    {        // Check if K is greater than N/2        if (K > N / 2)            return "NO";         // Check if N is even and K is odd        if (N % 2 == 0 && K % 2 == 1)            return "YES";         return "NO";    }     static void Main()    {        int N = 16;        int K = 3;        Console.WriteLine(IsPossibleN(N, K));    }}

## Javascript

 function isPossibleN(N, K) {    // Check if K is greater than N/2    if (K > Math.floor(N / 2))        return "NO";     // Check if N is even and K is odd    if (N % 2 === 0 && K % 2 === 1)        return "YES";     return "NO";} const N = 16;const K = 3;console.log(isPossibleN(N, K));

Output

YES

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