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Divide first N natural numbers into 3 equal sum subsets
  • Last Updated : 13 Sep, 2019

Given an integer N, the task is to check whether the elements from the range [1, N] can be divided into three non-empty equal sum subsets. If possible then print Yes else print No.

Examples:

Input: N = 5
Output: Yes
The possible subsets are {1, 4}, {2, 3} and {5}.
(1 + 4) = (2 + 3) = (5)

Input: N = 3
Output: No

Approach: There are two cases:



  1. If N ≤ 3: In this case, it is not possible to divide the elements in the subsets that satisfy the given condition. So, print No.
  2. If N > 3: In this case, it is only possible when the sum of all the elements of the range [1, N] is divisible by 3 which can be easily calculated as sum = (N * (N + 1)) / 2. Now, if sum % 3 = 0 then print Yes else print No.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function that returns true
// if the subsets are possible
bool possible(int n)
{
  
    // If n <= 3 then it is not possible
    // to divide the elements in three subsets
    // satisfying the given conditions
    if (n > 3) {
  
        // Sum of all the elements
        // in the range [1, n]
        int sum = (n * (n + 1)) / 2;
  
        // If the sum is divisible by 3
        // then it is possible
        if (sum % 3 == 0) {
            return true;
        }
    }
    return false;
}
  
// Driver code
int main()
{
    int n = 5;
  
    if (possible(n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.math.*;
  
class GFG
{
  
    // Function that returns true 
    // if the subsets are possible 
    public static boolean possible(int n) 
    
      
        // If n <= 3 then it is not possible 
        // to divide the elements in three subsets 
        // satisfying the given conditions 
        if (n > 3
        
      
            // Sum of all the elements 
            // in the range [1, n] 
            int sum = (n * (n + 1)) / 2
      
            // If the sum is divisible by 3 
            // then it is possible 
            if (sum % 3 == 0
            
                return true
            
        
        return false
    
  
    // Driver code 
    public static void main(String[] args) 
    
        int n = 5
  
        if (possible(n)) 
            System.out.println("Yes"); 
        else
            System.out.println("No");
    }
}
  
// This code is contributed by Naman_Garg

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Python3

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# Python3 implementation of the approach 
  
# Function that returns true 
# if the subsets are possible 
def possible(n) : 
  
    # If n <= 3 then it is not possible 
    # to divide the elements in three subsets 
    # satisfying the given conditions 
    if (n > 3) :
  
        # Sum of all the elements 
        # in the range [1, n] 
        sum = (n * (n + 1)) // 2
  
        # If the sum is divisible by 3 
        # then it is possible 
        if (sum % 3 == 0) :
            return True
      
    return False
  
# Driver code 
if __name__ == "__main__"
  
    n = 5
  
    if (possible(n)) :
        print("Yes"); 
    else :
        print("No"); 
          
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach 
using System;
  
class GFG
{
      
// Function that returns true 
// if the subsets are possible 
public static bool possible(int n) 
  
    // If n <= 3 then it is not possible 
    // to divide the elements in three subsets 
    // satisfying the given conditions 
    if (n > 3) 
    
  
        // Sum of all the elements 
        // in the range [1, n] 
        int sum = (n * (n + 1)) / 2; 
  
        // If the sum is divisible by 3 
        // then it is possible 
        if (sum % 3 == 0) 
        
            return true
        
    
    return false
  
// Driver code 
static public void Main ()
{
    int n = 5; 
  
    if (possible(n)) 
        Console.Write("Yes"); 
    else
        Console.Write("No");
}
}
  
// This code is contributed by ajit

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Output:

Yes

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