# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to check if a  ` `// given number is sum of  ` `// cubes of natural numbers. ` `#include ` `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; ` `} `

## Java

 `// 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); ` `    ``} ` `} `

## Python3

 `# 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) `

## C#

 `// 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. `

## PHP

 ` `

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++

 `// C++ program to check if a  ` `// given number is sum of  ` `// cubes of natural numbers. ` `#include ` `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; ` `} `

## Java

 `// 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. `

## Python3

 `# 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. `

## C#

 `// 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. `

## PHP

 ` 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. ` `?> `

Output :

```2
```

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