Find if given number is sum of first n natural numbers

Given a number s (1 <= s <= 1000000000). If this number is the sum of first n natural number then print n, otherwise print -1. 

Examples:

Input: s = 10
Output: n = 4
Explanation: 1 + 2 + 3 + 4 = 10

Input: s = 17
Output: n = -1
Explanation: 17 can’t be expressed as a sum of first n natural numbers as sum of first 5 natural numbers is 15 and sum of first 6 natural numbers is 21

Finding if given number is sum of first n natural numbers using Brute Force:

The idea is to keep adding numbers starting from 1 and if the sum of the numbers become equal to s, then return the count of numbers added so far. If the sum becomes greater than s, then return -1 as it is impossible to express s as sum of first n natural numbers.

Step by step approach:

• Maintain a variable sum = 0 to calculate the running sum of numbers.
• Iterate a loop from i = 1 to s
  • Add i to sum
  • If sum == s, return i
  • Else if sum > s, return -1

Below is the implementation of the above approach:

C++

// C++ program for above implementation
#include <iostream>
using namespace std;

// Function to find no. of elements
// to be added from 1 to get sum = s
int findS(int s)
{
    int sum = 0;

    // Start adding numbers from 1
    for (int i = 1; sum < s; i++) {
        sum += i;

        // If sum becomes equal to s
        // return i
        if (sum == s)
            return i;
    }

    return -1;
}

// Drivers code
int main()
{
    int s = 15;
    cout << findS(s);

    return 0;
}

Java

// Java program for above implementation
class GFG {

    // Function to find no. of elements
    // to be added from 1 to get sum = s
    static int findS(int s)
    {
        int sum = 0;

        // Start adding numbers from 1
        for (int i = 1; sum < s; i++) {
            sum += i;

            // If sum becomes equal to s
            // return i
            if (sum == s)
                return i;
        }

        return -1;
    }

    // Drivers code
    public static void main(String[] args)
    {

        int s = 15;
        System.out.println(findS(s));
    }
}

// This code is contributed by Azkia Anam.

Python3

# Python3 program to check if
# given number is sum of first n
# natural numbers

# Function to find no. of elements
# to be added from 1 to get sum = s


def findS(s):
    _sum = 0
    i = 1

    # Start adding numbers from 1
    while(_sum < s):
        _sum += i
        # If sum becomes equal to s
        # return i
        if _sum == s:
            return i
        i += 1

    return -1


# Driver code
s = 15
print(findS(s))

# This code is contributed by "Abhishek Sharma 44".

C#

// C# program for above implementation
using System;

class GFG {

    // Function to find no. of elements
    // to be added from 1 to get sum = s
    static int findS(int s)
    {
        int sum = 0;

        // Start adding numbers from 1
        for (int i = 1; sum < s; i++) {
            sum += i;

            // If sum becomes equal to s
            // return i
            if (sum == s)
                return i;
        }

        return -1;
    }

    // Drivers code
    public static void Main()
    {

        int s = 15;

        Console.WriteLine(findS(s));
    }
}

// This code is contributed by vt_m.

Javascript

<script>

// Javascript program for above implementation

// Function to find no. of elements
// to be added from 1 to get sum = s
    function findS(s) {
        var sum = 0;

        // Start adding numbers from 1
        for (i = 1; sum < s; i++) {
            sum += i;

            // If sum becomes equal to s
            // return i
            if (sum == s)
                return i;
        }

        return -1;
    }

    // Drivers code
    var s = 15;
    document.write(findS(s));

// This code is contributed by Rajput-Ji

</script>

PHP

<?php
// PHP program for above implementation

// Function to find no. of elements
// to be added from 1 to get sum = s
function findS($s)
{
    $sum = 0;

    // Start adding numbers from 1
    for ($i = 1; $sum < $s; $i++)
    {
        $sum += $i;

        // If sum becomes equal
        // to s return i
        if ($sum == $s)
            return $i;
    }

    return -1;
}

// Drivers code
$s = 15;
echo findS($s);

// This code is contributed by Sam007
?>

Output

5

Time Complexity: O(√s), where s is the number we need to check as the sum of first n natural numbers
Auxiliary Space: O(1)

Finding if given number is sum of first n natural numbers using Mathematical formula:

The idea is to use the formula of the sum of first N natural numbers to compute the value of the N. Below is the illustration:

⇾ 1 + 2 + 3 + …. N =  S
⇾ (N * (N + 1)) /  2 = S
⇾ N * (N + 1) = 2 * S
⇾ N²  + N  – 2 * S = 0

Therefore, Calculate the solution of the quadratic equation using Sridharacharya Formula and check if the solution is an integer or not. If Yes then the solution exists. Otherwise, the given number is not sum of first N natural number.

Step by step algorithm:

• Calculate the positive root of equation: N²  + N  – 2 * S = 0
• Check if the root is an integer or not
• Return root if it is an integer, else return -1.

Below is the implementation of the above approach:

C++

// C++ program of the above
// approach

#include <bits/stdc++.h>

#define ll long long

using namespace std;

// Function to check if the
// s is the sum of first N
// natural number
ll int isvalid(ll int s)
{
    // Solution of Quadratic Equation
    float k = (-1 + sqrt(1 + 8 * s)) / 2;

    // Condition to check if the
    // solution is a integer
    if (ceil(k) == floor(k))
        return k;
    else
        return -1;
}

// Driver Code
int main()
{
    int s = 15;

    // Function Call
    cout << isvalid(s);
    return 0;
}

Java

// Java program of the above
// approach
import java.util.*;

class GFG {

    // Function to check if the
    // s is the sum of first N
    // natural number
    public static int isvalid(int s)
    {

        // Solution of Quadratic Equation
        double k = (-1.0 + Math.sqrt(1 + 8 * s)) / 2;

        // Condition to check if the
        // solution is a integer
        if (Math.ceil(k) == Math.floor(k))
            return (int)k;
        else
            return -1;
    }

    // Driver code
    public static void main(String[] args)
    {
        int s = 15;

        // Function call
        System.out.print(isvalid(s));
    }
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program of the above
# approach
import math

# Function to check if the
# s is the sum of first N
# natural number


def isvalid(s):

    # Solution of Quadratic Equation
    k = (-1 + math.sqrt(1 + 8 * s)) / 2

    # Condition to check if the
    # solution is a integer
    if (math.ceil(k) == math.floor(k)):
        return int(k)
    else:
        return -1


# Driver Code
s = 15

# Function Call
print(isvalid(s))

# This code is contributed by vishu2908

C#

// C# program of the above
// approach
using System;
class GFG {

    // Function to check if the
    // s is the sum of first N
    // natural number
    public static int isvalid(int s)
    {
        // Solution of Quadratic Equation
        double k = (-1.0 + Math.Sqrt(1 + 8 * s)) / 2;

        // Condition to check if the
        // solution is a integer
        if (Math.Ceiling(k) == Math.Floor(k))
            return (int)k;
        else
            return -1;
    }

    // Driver code
    public static void Main(string[] args)
    {
        int s = 15;

        // Function call
        Console.Write(isvalid(s));
    }
}

// This code is contributed by Chitranayal

Javascript

<script>

// Javascript program of the above
// approach

// Function to check if the
// s is the sum of first N
// natural number
function isvalid(s)
{

    // Solution of Quadratic Equation
    let k = (-1.0 + Math.sqrt(1 + 8 * s)) / 2;

    // Condition to check if the
    // solution is a integer
    if (Math.ceil(k) == Math.floor(k))
        return k;
    else
      return -1;
}
// Driver code

        let s = 15;

    // Function call
    document.write(isvalid(s));

</script>

Output

5

Time Complexity: O(log(s)) because it is using sqrt() function
Auxiliary Space: O(1)