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 = 14Input : 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 numbersint 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 codeint 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 numbersint 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 codeint 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 numbersdef 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 codes = 13n = 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 numbersfunction 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.
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