Squared triangular number (Sum of cubes)

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

First few Squared triangular number are 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, …

Examples :



Input : 9
Output : 2
Explanation : The given number is
sum of cubes of first 2 natural
numbers.  1*1*1 + 2*2*2 = 9

Input : 13
Output : -1

A simple solution is to one by one add cubes of natural numbers. If current sum becomes same as given number, then we return count of natural numbers added so far. Else we return -1.

C++

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// C++ program to check if a 
// given number is sum of 
// cubes of natural numbers.
#include <iostream>
using namespace std;
  
// Function to find if 
// the given number is 
// sum of the cubes of 
// first n natural numbers
int findS(int s)
{
    int sum = 0;
  
    // Start adding cubes of 
    // the numbers from 1
    for (int n = 1; sum < s; n++) 
    {
        sum += n * n * n;
  
        // If sum becomes equal to s
        // return n
        if (sum == s)
            return n;
    }
  
    return -1;
}
  
// Driver code
int main()
{
    int s = 9;
    int n = findS(s);
    n == -1 ? cout << "-1" : cout << n;
  
    return 0;
}

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Java

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// Java program to check if 
// a given number is sum of 
// cubes of natural numbers.
class GFG
{
  
    // Function to find if 
    // the given number is
    // sum of the cubes of 
    // first n natural numbers
    static int findS(int s)
    {
        int sum = 0;
  
        // Start adding cubes of 
        // the numbers from 1
        for (int n = 1; sum < s; n++) 
        {
            sum += n * 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 = 9;
        int n = findS(s);
        if (n == -1)
            System.out.println("-1");
        else
            System.out.println(n);
    }
}

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Python3

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# Python3 program to find
# if the given number is 
# sum of the cubes of first 
# n natural numbers
  
# Function to find if the 
# given number is sum of 
# the cubes of first n 
# natural numbers
def findS (s):
    _sum = 0
    n = 1
      
    # Start adding cubes of
    # the numbers from 1
    while(_sum < s):
        _sum += n * n * n
        n += 1
    n-= 1
      
    # If sum becomes equal to s
    # return n
    if _sum == s:
        return n
    return -1
  
# Driver code
s = 9
n = findS (s)
if n == -1:
    print("-1")
else:
    print(n)

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C#

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// C# program to check if a 
// given number is sum of 
// cubes of natural numbers.
using System;
  
class GFG
{
      
    // Function to find if the 
    // given number is sum of
    // the cubes of first n 
    // natural numbers
    public static int findS(int s)
    {
        int sum = 0;
  
        // Start adding cubes of 
        // the numbers from 1
        for (int n = 1; sum < s; n++) 
        {
            sum += n * n * n;
  
            // If sum becomes equal to s
            // return n
            if (sum == s)
                return n;
        }
  
        return -1;
    }
      
// Driver code
static public void Main (string []args)
{
      
    int s = 9;
    int n = findS(s);
    if (n == -1)
        Console.WriteLine("-1");
    else
        Console.WriteLine(n);
}
}
  
// This code is contributed by Ajit.

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PHP

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<?php
// PHP program to check if 
// a given number is sum of 
// cubes of natural numbers.
  
// Function to find if the 
// given number is sum of 
// the cubes of first n 
// natural numbers
function findS($s)
{
    $sum = 0;
  
    // Start adding cubes of 
    // the numbers from 1
    for ($n = 1; $sum < $s; $n++) 
    {
        $sum += $n * $n * $n;
  
        // If sum becomes equal to s
        // return n
        if ($sum == $s)
            return $n;
    }
  
    return -1;
}
  
// Driver code
$s = 9;
$n = findS($s);
if($n == -1)
    echo("-1");
else
    echo($n);
  
// This code is contributed by Ajit.
?>

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Output:

2

An efficient solution is based on the formula [n(n+1)/2]2 for sum of first n cubes. We can see that all numbers are squares.
1) Check if given number is perfect square.
2) Check if square root is triangular (Please see method 2 of triangular numbers for this)

C++

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// C++ program to check if a 
// given number is sum of 
// cubes of natural numbers.
#include <bits/stdc++.h>
using namespace std;
  
// Returns root of n(n+1)/2 = num
// if num is triangular (or integer
// root exists). Else returns -1.
int isTriangular(int num)
{
    if (num < 0)
        return false;
  
    // Considering the equation 
    // n*(n+1)/2 = num. The equation 
    // is : a(n^2) + bn + c = 0";
    int c = (-2 * num);
    int b = 1, a = 1;
    int d = (b * b) - (4 * a * c);
  
    if (d < 0)
        return -1;
  
    // Find roots of equation
    float root1 = ( -b + sqrt(d)) / (2 * a);
    float root2 = ( -b - sqrt(d)) / (2 * a);
  
    // checking if root1 is natural
    if (root1 > 0 && floor(root1) == root1)
        return root1;
  
    // checking if root2 is natural
    if (root2 > 0 && floor(root2) == root2)
        return root2;
  
    return -1;
}
  
// Returns square root of x if it is 
// perfect square. Else returns -1.
int isPerfectSquare(long double x)
// Find floating point value of 
// square root of x.
long double sr = sqrt(x);
  
// If square root is an integer
if ((sr - floor(sr)) == 0)
    return floor(sr);
else
    return -1;
}
  
// Function to find if the given number
// is sum of the cubes of first n
// natural numbers
int findS(int s)
{
    int sr = isPerfectSquare(s);
    if (sr == -1)
    return -1;
    return isTriangular(sr);
  
// Driver code
int main()
{
    int s = 9;
    int n = findS(s);
    n == -1 ? cout << "-1" : cout << n;
  
    return 0;
}

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Java

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// Java program to check 
// if a given number is 
// sum of cubes of natural 
// numbers. 
// import java.Math.*;
class GFG
{
      
// Returns root of n(n+1)/2 = num
// if num is triangular (or 
// integer root exists). Else 
// returns -1.
public static int isTriangular(int num)
{
    if (num < 0)
        return 0;
  
    // Considering the equation 
    // n*(n+1)/2 = num. The equation 
    // is : a(n^2) + bn + c = 0";
    int c = (-2 * num);
    int b = 1, a = 1;
    int d = (b * b) - 
            (4 * a * c);
  
    if (d < 0)
        return -1;
  
    // Find roots of equation
    double root1 = (-b + 
           Math.sqrt(d)) / (2 * a);
    double root2 = (-b - 
           Math.sqrt(d)) / (2 * a);
  
    // checking if root1 is natural
    if ((int)(root1) > 0 && 
        (int)(Math.floor(root1)) == 
                   (int)(root1))
        return (int)(root1);
  
    // checking if 
    // root2 is natural
    if ((int)(root2) > 0 && 
        (int)(Math.floor(root2)) == 
                   (int)(root2))
        return (int)(root2);
  
    return -1;
}
  
// Returns square root 
// of x if it is perfect 
// square. Else returns -1.
static int isPerfectSquare(double x)
      
// Find floating point 
// value of square root of x.
double sr = Math.sqrt(x);
  
// If square root
// is an integer
if ((sr - Math.floor(sr)) == 0)
    return (int)(Math.floor(sr));
else
    return -1;
}
  
// Function to find if the 
// given number is sum of 
// the cubes of first n
// natural numbers
static int findS(int s)
{
    int sr = isPerfectSquare(s);
    if (sr == -1)
    return -1;
    return isTriangular(sr);
  
// Driver code
public static void main(String[] args)
{
    int s = 9;
    int n = findS(s);
    if(n == -1)
    System.out.println("-1");
    else
    System.out.println(n);
}
}
  
// This code is contributed
// by mits.

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Python3

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# Python3 program to check 
# if a given number is sum of 
# cubes of natural numbers.
import math
  
# Returns root of n(n+1)/2 = num
# if num is triangular (or integer
# root exists). Else returns -1.
def isTriangular(num):
    if (num < 0):
        return False;
  
    # Considering the equation 
    # n*(n+1)/2 = num. The equation 
    # is : a(n^2) + bn + c = 0";
    c = (-2 * num);
    b = 1;
    a = 1;
    d = (b * b) - (4 * a * c);
  
    if (d < 0):
        return -1;
  
    # Find roots of equation
    root1 = (-b + math.sqrt(d)) // (2 * a);
    root2 = (-b - math.sqrt(d)) // (2 * a);
  
    # checking if root1 is natural
    if (root1 > 0 and 
        math.floor(root1) == root1):
        return root1;
  
    # checking if root2 is natural
    if (root2 > 0 and 
        math.floor(root2) == root2):
        return root2;
  
    return -1;
  
# Returns square root of 
# x if it is perfect square.
# Else returns -1.
def isPerfectSquare(x):
      
    # Find floating point value
    # of square root of x.
    sr = math.sqrt(x);
      
    # If square root is an integer
    if ((sr - math.floor(sr)) == 0):
        return math.floor(sr);
    else:
        return -1;
  
# Function to find if the given
# number is sum of the cubes of
# first n natural numbers
def findS(s):
    sr = isPerfectSquare(s);
    if (sr == -1):
        return -1;
    return int(isTriangular(sr));
  
# Driver code
s = 9;
n = findS(s);
if(n == -1):
    print("-1");
else:
    print(n);
      
# This code is contributed by mits.

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C#

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// C# program to check if a 
// given number is sum of 
// cubes of natural numbers.
using System;
  
class GFG
{
      
// Returns root of n(n+1)/2 = num
// if num is triangular (or integer
// root exists). Else returns -1.
static int isTriangular(int num)
{
    if (num < 0)
        return 0;
  
    // Considering the equation 
    // n*(n+1)/2 = num. The equation 
    // is : a(n^2) + bn + c = 0";
    int c = (-2 * num);
    int b = 1, a = 1;
    int d = (b * b) - 
            (4 * a * c);
  
    if (d < 0)
        return -1;
  
    // Find roots of equation
    double root1 = (-b + Math.Sqrt(d)) / (2 * a);
    double root2 = (-b - Math.Sqrt(d)) / (2 * a);
  
    // checking if root1 is natural
    if ((int)(root1) > 0 && 
        (int)(Math.Floor(root1)) == (int)(root1))
        return (int)(root1);
  
    // checking if root2 is natural
    if ((int)(root2) > 0 && 
        (int)(Math.Floor(root2)) == (int)(root2))
        return (int)(root2);
  
    return -1;
}
  
// Returns square root of x 
// if it is perfect square. 
// Else returns -1.
static int isPerfectSquare(double x)
      
// Find floating point 
// value of square root of x.
double sr = Math.Sqrt(x);
  
// If square root
// is an integer
if ((sr - Math.Floor(sr)) == 0)
    return (int)(Math.Floor(sr));
else
    return -1;
}
  
// Function to find if the 
// given number is sum of 
// the cubes of first n
// natural numbers
static int findS(int s)
{
    int sr = isPerfectSquare(s);
    if (sr == -1)
    return -1;
    return isTriangular(sr);
  
// Driver code
public static void Main()
{
    int s = 9;
    int n = findS(s);
    if(n == -1)
    Console.Write("-1");
    else
    Console.Write(n);
}
}
  
// This code is contributed by mits.

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PHP

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<?php
// PHP program to check if a 
// given number is sum of 
// cubes of natural numbers.
  
// Returns root of n(n+1)/2 = num
// if num is triangular (or integer
// root exists). Else returns -1.
function isTriangular($num)
{
    if ($num < 0)
        return false;
  
    // Considering the equation 
    // n*(n+1)/2 = num. The equation 
    // is : a(n^2) + bn + c = 0";
    $c = (-2 * $num);
    $b = 1;
    $a = 1;
    $d = ($b * $b) - (4 * $a * $c);
  
    if ($d < 0)
        return -1;
  
    // Find roots of equation
    $root1 = (-$b + sqrt($d)) / 
                      (2 * $a);
    $root2 = (-$b - sqrt($d)) / 
                      (2 * $a);
  
    // checking if root1 is natural
    if ($root1 > 0 &&
            floor($root1) == $root1)
        return $root1;
  
    // checking if root2 is natural
    if ($root2 > 0 && 
            floor($root2) == $root2)
        return $root2;
  
    return -1;
}
  
// Returns square root of  
// x if it is perfect square.
// Else returns -1.
function isPerfectSquare($x)
      
// Find floating point value 
// of square root of x.
$sr = sqrt($x);
  
// If square root is an integer
if (($sr - floor($sr)) == 0)
    return floor($sr);
else
    return -1;
}
  
// Function to find if the given
// number is sum of the cubes of
// first n natural numbers
function findS($s)
{
    $sr = isPerfectSquare($s);
    if ($sr == -1)
    return -1;
    return isTriangular($sr);
  
// Driver code
$s = 9;
$n = findS($s);
if($n == -1)
echo "-1";
else
echo $n;
      
// This code is contributed by mits.
?>

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Output :

2


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