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++
// 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 numbersint 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 codeint main(){ int s = 9; int n = findS(s); n == -1 ? cout << "-1" : cout << n; return 0;} |
C
// C program to check if a // given number is sum of // cubes of natural numbers.#include <stdio.h>// Function to find if // the given number is // sum of the cubes of // first n natural numbersint 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 codeint main(){ int s = 9; 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 // 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 numbersdef 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 codes = 9n = 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 codestatic 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
<?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 numbersfunction 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.?> |
Javascript
<script>// Javascript 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 numbersfunction findS(s){ let sum = 0; // Start adding cubes of // the numbers from 1 for (let 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 let s = 9; let n = findS(s); n == -1 ? document.write("-1") :document.write(n);// This code is contributed by aashish1995</script> |
Output:
2
Time Complexity: O(n)
Auxiliary Space: O(1)
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 <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 integerif ((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 numbersint findS(int s){ int sr = isPerfectSquare(s); if (sr == -1) return -1; return isTriangular(sr);} // Driver codeint main(){ int s = 9; int n = findS(s); n == -1 ? cout << "-1" : cout << n; return 0;} |
C
// C program to check if a // given number is sum of // cubes of natural numbers.#include <stdio.h>#include <stdbool.h>#include <math.h>// 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 integerif ((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 numbersint findS(int s){ int sr = isPerfectSquare(s); if (sr == -1) return -1; return isTriangular(sr);} // Driver codeint main(){ int s = 9; 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 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 integerif ((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 numbersstatic int findS(int s){ int sr = isPerfectSquare(s); if (sr == -1) return -1; return isTriangular(sr);} // Driver codepublic 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 numbersdef findS(s): sr = isPerfectSquare(s); if (sr == -1): return -1; return int(isTriangular(sr));# Driver codes = 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 integerif ((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 numbersstatic int findS(int s){ int sr = isPerfectSquare(s); if (sr == -1) return -1; return isTriangular(sr);} // Driver codepublic 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
<?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 integerif (($sr - floor($sr)) == 0) return floor($sr);else return -1;}// Function to find if the given// number is sum of the cubes of// first n natural numbersfunction findS($s){ $sr = isPerfectSquare($s); if ($sr == -1) return -1; return isTriangular($sr);} // Driver code$s = 9;$n = findS($s);if($n == -1)echo "-1";elseecho $n; // This code is contributed by mits.?> |
Javascript
<script>// javascript 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 0; // Considering the equation // n*(n+1)/2 = num. The equation // is : a(n^2) + bn + c = 0"; var c = (-2 * num); var b = 1, a = 1; var d = (b * b) - (4 * a * c); if (d < 0) return -1; // Find roots of equation var root1 = (-b + Math.sqrt(d)) / (2 * a); var root2 = (-b - Math.sqrt(d)) / (2 * a); // checking if root1 is natural if (parseInt( (root1)) > 0 && parseInt( (Math.floor(root1))) == parseInt( (root1))) return parseInt(root1); // checking if // root2 is natural if (parseInt( (root2)) > 0 && parseInt( (Math.floor(root2))) == parseInt( (root2))) return parseInt( (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. var sr = Math.sqrt(x); // If square root // is an integer if ((sr - Math.floor(sr)) == 0) return parseInt( (Math.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) { var sr = isPerfectSquare(s); if (sr == -1) return -1; return isTriangular(sr); } // Driver code var s = 9; var n = findS(s); if (n == -1) document.write("-1"); else document.write(n);// This code is contributed by Rajput-Ji. </script> |
Output :
2
Time Complexity: O(logn)
Auxiliary Space: O(1)
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