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Square pyramidal number (Sum of Squares)

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A Square pyramidal number represents sum of squares of first natural numbers. First few Square pyramidal numbers are 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, …
Geometrically these numbers represent number of spheres to be stacked to form a pyramid with square base. Please see this Wiki image for more clarity.
Given a number s (1 <= s <= 1000000000). If s is sum of the squares of the first n natural numbers then print n, otherwise print -1.
Examples : 
 

Input : 14
Output : 3
Explanation : 1*1 + 2*2 + 3*3 = 14

Input : 26
Output : -1

 

A simple solution is to run through all numbers starting from 1, compute current sum. If current sum is equal to given sum, then we return true, else false.
 

C++




// C++ program to check if a
// given number is sum of
// squares of natural numbers.
#include <iostream>
using namespace std;
 
// Function to find if the
// given number is sum of
// the squares of first n
// natural numbers
int findS(int s)
{
    int sum = 0;
 
    // Start adding squares of
    // the numbers from 1
    for (int n = 1; sum < s; n++)
    {
        sum += n * n;
 
        // If sum becomes equal to s
        // return n
        if (sum == s)
            return n;
    }
 
    return -1;
}
 
// Drivers code
int main()
{
    int s = 13;
    int n = findS(s);
    n == -1 ? cout << "-1" : cout << n;
 
    return 0;
}


C




// C program to check if a
// given number is sum of
// squares of natural numbers.
#include <stdio.h>
 
// Function to find if the
// given number is sum of
// the squares of first n
// natural numbers
int findS(int s)
{
    int sum = 0;
 
    // Start adding squares of
    // the numbers from 1
    for (int n = 1; sum < s; n++)
    {
        sum += n * n;
 
        // If sum becomes equal to s
        // return n
        if (sum == s)
            return n;
    }
 
    return -1;
}
 
// Drivers code
int main()
{
    int s = 13;
    int n = findS(s);
     
    n == -1 ? printf("-1") : printf("%d",n);
 
    return 0;
}
 
// This code is contributed by kothavvsaakash.


Java




// Java program to check if a
// given number is sum of
// squares of natural numbers.
class GFG
{
 
    // Function to find if the
    // given number is sum of
    // the squares of first
    // n natural numbers
    static int findS(int s)
    {
        int sum = 0;
 
        // Start adding squares of
        // the numbers from 1
        for (int n = 1; sum < s; n++)
        {
            sum += n * n;
 
            // If sum becomes equal to s
            // return n
            if (sum == s)
                return n;
        }
 
        return -1;
    }
 
    // Drivers code
    public static void main(String[] args)
    {
 
        int s = 13;
        int n = findS(s);
        if (n == -1)
            System.out.println("-1");
        else
            System.out.println(n);
    }
}


Python3




# Python3 program to find if
# the given number is sum of
# the squares of first
# n natural numbers
 
# Function to find if the given
# number is sum of the squares
# of first n natural numbers
def findS (s):
    _sum = 0
    n = 1
     
    # Start adding squares of
    # the numbers from 1
    while(_sum < s):
        _sum += n * n
        n+= 1
    n-= 1
     
    # If sum becomes equal to s
    # return n
    if _sum == s:
        return n
    return -1
 
# Driver code
s = 13
n = findS (s)
if n == -1:
    print("-1")
else:
    print(n)


C#




// C# program to check if a given
// number is sum of squares of
// natural numbers.
using System;
 
class GFG
{
     
    // Function to find if the given
    // number is sum of the squares
    // of first n natural numbers
    static int findS(int s)
    {
        int sum = 0;
     
        // Start adding squares of
        // the numbers from 1
        for (int n = 1; sum < s; n++)
        {
            sum += n * n;
     
            // If sum becomes equal
            // to s return n
            if (sum == s)
                return n;
        }
     
        return -1;
    }
     
    // Drivers code
    public static void Main()
    {
        int s = 13;
         
        int n = findS(s);
         
        if(n == -1)
            Console.Write("-1") ;
        else
            Console.Write(n);
    }
}
// This code is contribute by
// Smitha Dinesh Semwal


PHP




<?php
// PHP program to check if a
// given number is sum of
// squares of natural numbers.
 
// Function to find if the given number
// is sum of the squares of first n
// natural numbers
function findS($s)
{
    $sum = 0;
 
    // Start adding squares of
    // the numbers from 1
    for ($n = 1; $sum < $s; $n++)
    {
        $sum += $n * $n;
 
        // If sum becomes equal to s
        // return n
        if ($sum == $s)
            return $n;
    }
 
    return -1;
}
 
// Drivers code
$s = 13;
$n = findS($s);
if($n == -1)
    echo("-1");
else
    echo($n);
 
// This code is contributed by Ajit.
?>


Javascript




<script>
 
// JavaScript to check if a
// given number is sum of
// squares of natural numbers.
 
    // Function to find if the
    // given number is sum of
    // the squares of first
    // n natural numbers
    function findS(s)
    {
        let sum = 0;
   
        // Start adding squares of
        // the numbers from 1
        for (let n = 1; sum < s; n++)
        {
            sum += n * n;
   
            // If sum becomes equal to s
            // return n
            if (sum == s)
                return n;
        }
   
        return -1;
    }
 
// Driver code
        let s = 13;
        let n = findS(s);
        if (n == -1)
            document.write("-1");
        else
            document.write(n);
 
// This code is contributed by souravghosh0416.
</script>


OUTPUT : 
 

-1

Time complexity: O(s) for given input s
Space complexity: O(1)

An alternate solution is to use Newton Raphson Method
We know sum of squares of first n natural numbers is n * (n + 1) * (2*n + 1) / 6.
We can write solutions as
k * (k + 1) * (2*k + 1) / 6 = s
k * (k + 1) * (2*k + 1) – 6s = 0
We can find roots of above cubic equation using Newton Raphson Method, then check if root is integer or not.
 



Last Updated : 20 Sep, 2022
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