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Check if K distinct positive even integers with given sum exists or not
  • Last Updated : 08 Feb, 2021

Given two integers N and K, the task is to find if there exists 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 <bits/stdc++.h>
using namespace std;
 
// Function to find if the sum
// of K even integers equals N
void 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 then equal to N
    if (sum <= N) {
        cout << "YES" << endl;
    }
 
    // Otherwise
    else {
        cout << "No" << endl;
    }
 
    return;
}
 
// Driver Code
int main()
{
 
    int N = 16;
    int K = 3;
 
    isPossibleN(N, K);
 
    return 0;
}

Java




// Java program for the above approach
import 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 then equal to N
    if (sum <= N) {
        System.out.println("YES");
    }
 
    // Otherwise
    else {
        System.out.println("No");
    }
}
 
// Driver Code
public 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 N
def 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 then equal to N
    if (sum <= N) :
        print("YES")
     
  
    # Otherwise
    else :
        print("NO")
  
    return
 
  
# Driver Code
 
N = 16
K = 3
  
isPossibleN(N, K)

C#




// C# implementation of the
// above approach
using 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 then 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.

  

Output: 
YES

 

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

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