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Check if a given number can be expressed as pair-sum of sum of first X natural numbers
  • Difficulty Level : Medium
  • Last Updated : 31 Mar, 2021

Given an integer N, the task is to check if N is the sum of a pair of integers which can be expressed as the sum of first X natural numbers, where X can be any positive integer. If satisfies the required condition. Print “YES”. Otherwise, print “NO”.

Examples:

Input: N = 25
Output: YES
Explanation:
=> 10 + 15 = 25
Since 10 and 15 are the sum of first 4 and 5 natural numbers respectively, the answer is YES.

Input: N = 512
Output: NO

Approach: The idea is to choose a sum of natural numbers M which is less than equal to N and check if M and N – M are the sums of the sequence of the first few natural numbers. Follow the steps below to solve the problem:



  • Iterate over a loop to calculate the sum of K natural numbers:

 Sum of K natural numbers = K * (K + 1) / 2 

  • Then, calculate the remaining sum and check if the sum is the sum by the following equation:

 Y = N – Sum of K Natural number 
=> Y = N – (K * (K + 1) / 2) 

  • Check if the number calculated above satisfies the required condition by calculating the square root of the twice of the number and check if the product of consecutive numbers is equal to the twice of the number.

 M * (M + 1) == 2 * Y, where M = √ (2 * Y) 

  • If the above condition is satisfied, print “YES”. Otherwise, print “NO”.

Below is the implementation of the above approach:

C++




// C++ program of the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to check if the number
// is pair-sum of sum of first X
// natural numbers
void checkSumOfNatural(int n)
{
    int i = 1;
    bool flag = false;
     
    // Check if the given number
    // is sum of pair of special numbers
    while (i * (i + 1) < n * 2)
    {
         
        // X is the sum of first
        // i natural numbers
        int X = i * (i + 1);
         
        // t = 2 * Y
        int t = n * 2 - X;
        int k = sqrt(t);
         
        // Condition to check if
        // Y is a special number
        if (k * (k + 1) == t)
        {
            flag = true;
            break;
        }
        i += 1;
    }
     
    if (flag)
        cout << "YES";
    else
        cout << "NO";
}
 
// Driver Code
int main()
{
    int n = 25;
     
    // Function call
    checkSumOfNatural(n);
 
    return 0;
}
 
// This code is contributed by rutvik_56

Java




// Java program of the above approach
import java.util.*;
import java.lang.*;
 
class GFG{
 
// Function to check if the number
// is pair-sum of sum of first X
// natural numbers
static void checkSumOfNatural(int n)
{
    int i = 1;
    boolean flag = false;
     
    // Check if the given number
    // is sum of pair of special numbers
    while (i * (i + 1) < n * 2)
    {
         
        // X is the sum of first
        // i natural numbers
        int X = i * (i + 1);
         
        // t = 2 * Y
        int t = n * 2 - X;
        int k = (int)Math.sqrt(t);
         
        // Condition to check if
        // Y is a special number
        if(k * (k + 1) == t)
        {
            flag = true;
            break;
        }
        i += 1;
    }
     
    if (flag)
        System.out.println("YES");
    else
        System.out.println("NO");
}
 
// Driver Code
public static void main (String[] args)
{
    int n = 25;
     
    // Function call
    checkSumOfNatural(n);
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program of the
# above approach
 
import math
 
# Function to check if the number
# is pair-sum of sum of first X
# natural numbers
def checkSumOfNatural(n):
    i = 1
    flag = False
     
    # Check if the given number
    # is sum of pair of special numbers
    while i*(i + 1) < n * 2:
         
        # X is the sum of first
        # i natural numbers
        X = i*(i + 1)
         
        # t = 2 * Y
        t = n * 2 - X
        k = int(math.sqrt(t))
         
        # Condition to check if
        # Y is a special number
        if k*(k + 1) == t:
            flag = True
            break
        i += 1
     
    if flag:
        print('YES')
    else:
        print('NO')
 
# Driver Code       
if __name__ == "__main__":
    n = 25
     
    # Function Call
    checkSumOfNatural(n)

C#




// C# program of
// the above approach
using System;
class GFG{
 
// Function to check if the number
// is pair-sum of sum of first X
// natural numbers
static void checkSumOfNatural(int n)
{
  int i = 1;
  bool flag = false;
 
  // Check if the given number
  // is sum of pair of special numbers
  while (i * (i + 1) < n * 2)
  {
    // X is the sum of first
    // i natural numbers
    int X = i * (i + 1);
 
    // t = 2 * Y
    int t = n * 2 - X;
    int k = (int)Math.Sqrt(t);
 
    // Condition to check if
    // Y is a special number
    if(k * (k + 1) == t)
    {
      flag = true;
      break;
    }
    i += 1;
  }
 
  if (flag)
    Console.WriteLine("YES");
  else
    Console.WriteLine("NO");
}
 
// Driver Code
public static void Main(String[] args)
{
  int n = 25;
 
  // Function call
  checkSumOfNatural(n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
// javascript program of the above approach// Function to check if the number
// is pair-sum of sum of first X
// natural numbers
function checkSumOfNatural(n)
{
    var i = 1;
    var flag = false;
     
    // Check if the given number
    // is sum of pair of special numbers
    while (i * (i + 1) < n * 2)
    {
         
        // X is the sum of first
        // i natural numbers
        var X = i * (i + 1);
         
        // t = 2 * Y
        var t = n * 2 - X;
        var k = parseInt(Math.sqrt(t));
         
        // Condition to check if
        // Y is a special number
        if(k * (k + 1) == t)
        {
            flag = true;
            break;
        }
        i += 1;
    }
     
    if (flag)
        document.write("YES");
    else
        document.write("NO");
}
 
// Driver Code
var n = 25;
     
// Function call
checkSumOfNatural(n);
 
// This code is contributed by Princi Singh
</script>
Output: 
YES

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

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