Possible two sets from first N natural numbers difference of sums as D

Given N and D, find if it is possible to make two sets from first N natural numbers such that the difference between the sum of 2 sets(individually) is D.

Examples :

Input  : 5 7
Output : yes
Explanation: Keeping 1 and 3 in one set,
and 2, 4 and 5 are in other set.
Sum of set 1 = 4
Sum of set 2 = 11 
So, the difference D = 7 
Which is the required difference

Input  : 4 5
Output : no

Approach :

Let s1 and s2 be the two sets.
Here we know that
sum(s1) + sum(s2) = N*(N+1)/2 and
sum(s1) – sum(s2) = D

Adding above 2 equations, we get
2*sum(s1) = N*(N+1)/2 + D

If sum(S1) and sum(S2) are integers, then only we can split the first N natural numbers into two sets. For that N*(N+1)/2 + D must be an even number.

C++

// C++ program for implementing  
// above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function returns true if it is  
// possible to split into two 
// sets otherwise returns false 
bool check(int N, int D) 
{     
    int temp = (N * (N + 1)) / 2 + D; 
    return (temp % 2 == 0); 
} 
  
// Driver code 
int main() 
{ 
    int N = 5; 
    int M = 7; 
    if (check(N, M)) 
        cout << "yes"; 
    else
        cout << "no"; 
  
    return 0; 
} 

Java

// Java program for implementing  
// above approach 
  
class GFG 
{ 
      
    // Function returns true if it is  
    // possible to split into two 
    // sets otherwise returns false 
    static boolean check(int N, int D) 
    {  
        int temp = (N * (N + 1)) / 2 + D; 
        return (temp % 2 == 0); 
    } 
      
    // Driver code 
    static public void main (String args[]) 
    { 
        int N = 5; 
        int M = 7; 
        if (check(N, M)) 
            System.out.println("yes"); 
        else
            System.out.println("no"); 
    } 
} 
  
// This code is contributed by Smitha. 

Python3

# Python program for implementing 
# above approach 
  
# Function returns true if it is 
# possible to split into two 
# sets otherwise returns false 
def check(N, D): 
    temp = N * (N + 1) // 2 + D 
    return (bool(temp % 2 == 0)) 
  
# Driver code 
N = 5
M = 7
if check(N, M): 
    print("yes") 
else: 
    print("no") 
  
# This code is contributed by Shrikant13. 

C#

// C# program for implementing  
// above approach 
using System; 
  
class GFG 
{ 
      
    // Function returns true if it is  
    // possible to split into two 
    // sets otherwise returns false 
    static bool check(int N, int D) 
    {  
        int temp = (N * (N + 1)) / 2 + D; 
        return (temp % 2 == 0); 
    } 
      
    // Driver code 
    static public void Main () 
    { 
        int N = 5; 
        int M = 7; 
        if (check(N, M)) 
            Console.Write("yes"); 
        else
            Console.Write("no"); 
    } 
} 
  
// This code is contributed by Ajit. 

PHP

<?php 
// PHP program for implementing  
// above approach 
  
// Function returns true if it is  
// possible to split into two 
// sets otherwise returns false 
function check($N, $D) 
{  
    $temp = ($N * ($N + 1)) / 2 + $D; 
    return ($temp % 2 == 0); 
} 
  
// Driver code 
$N = 5; 
$M = 7; 
if (check($N, $M)) 
    echo("yes"); 
else
    echo("no"); 
  
// This code is contributed by Ajit. 

OUTPUT :

yes

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

C++

Java

Python3

C#

PHP

Recommended Posts: