Given an integer N, the task is to check if this number can be represented as the sum of two consecutive perfect cubes or not.
Examples:
Input: N = 35
Output: Yes
Explanation:
Since, 35 = 23 + 33, therefore the required answer is Yes.Input: N = 14
Output: No
Naive Approach: The simplest approach to solve the problem is to iterate from 1 to cube root of N and check if the sum of perfect cubes of any two consecutive numbers is equal to N or not. If found to be true, print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ Program of the// above approach#include <bits/stdc++.h>using namespace std;// Function to check if a number// can be expressed as the sum of// cubes of two consecutive numbersbool isCubeSum(int n){ for (int i = 1; i * i * i <= n; i++) { if (i * i * i + (i + 1) * (i + 1) * (i + 1) == n) return true; } return false;}// Driver Codeint main(){ int n = 35; if (isCubeSum(n)) cout << "Yes"; else cout << "No";} |
Java
// Java program of the// above approachimport java.util.*;class GFG{// Function to check if a number// can be expressed as the sum of// cubes of two consecutive numbersstatic boolean isCubeSum(int n){ for(int i = 1; i * i * i <= n; i++) { if (i * i * i + (i + 1) * (i + 1) * (i + 1) == n) return true; } return false;}// Driver Codepublic static void main(String[] args){ int n = 35; if (isCubeSum(n)) System.out.print("Yes"); else System.out.print("No");}}// This code is contributed by Amit Katiyar |
Python3
# Python3 program of the# above approach# Function to check if a number# can be expressed as the sum of# cubes of two consecutive numbersdef isCubeSum(n): for i in range(1, int(pow(n, 1 / 3)) + 1): if (i * i * i + (i + 1) * (i + 1) * (i + 1) == n): return True; return False;# Driver Codeif __name__ == '__main__': n = 35; if (isCubeSum(n)): print("Yes"); else: print("No");# This code is contributed by Amit Katiyar |
C#
// C# program of the// above approachusing System;class GFG{// Function to check if a number// can be expressed as the sum of// cubes of two consecutive numbersstatic bool isCubeSum(int n){ for(int i = 1; i * i * i <= n; i++) { if (i * i * i + (i + 1) * (i + 1) * (i + 1) == n) return true; } return false;}// Driver Codepublic static void Main(String[] args){ int n = 35; if (isCubeSum(n)) Console.Write("Yes"); else Console.Write("No");}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript Program of the// above approach// Function to check if a number// can be expressed as the sum of// cubes of two consecutive numbersfunction isCubeSum(n){ for (var i = 1; i * i * i <= n; i++) { if (i * i * i + (i + 1) * (i + 1) * (i + 1) == n) return true; } return false;}// Driver Codevar n = 35;if (isCubeSum(n)) document.write("Yes");else document.write("No");</script> |
Output:
Yes
Time Complexity: O(N1/3)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following observations:
- A number can be represented as the sum of the perfect cube of two consecutive numbers if the sum of the cube root of both consecutive numbers is equal to N.
- This can be checked by the formula:
![\lfloor \sqrt[3]{N} - 1 \rfloor ^3 + \lfloor \sqrt[3]{N} \rfloor^3](https://neveropen.co.za/wp-content/uploads/2023/10/wardslaus_quicklatex.com-59e74925b276ff644cc733957260e587_l3.png "Rendered by QuickLaTeX.com")
- For example, if N = 35, then check if the equation below is equal to N or not:
![\lfloor \sqrt[3]{35} - 1 \rfloor ^3 + \lfloor \sqrt[3]{35} \rfloor^3](https://neveropen.co.za/wp-content/uploads/2023/10/wardslaus_rubhabha_quicklatex.com-9c81706accdbadd9c5df0e455830d13a_l3.png "Rendered by QuickLaTeX.com")
Below is the implementation of the above approach:
C++
// C++ Program to// implement above approach#include <bits/stdc++.h>using namespace std;// Function to check that a number// is the sum of cubes of 2// consecutive numbers or notbool isSumCube(int N){ int a = cbrt(N); int b = a - 1; // Condition to check if a // number is the sum of cubes of 2 // consecutive numbers or not return ((a * a * a + b * b * b) == N);}// Driver Codeint main(){ int i = 35; // Function call if (isSumCube(i)) { cout << "Yes"; } else { cout << "No"; } return 0;} |
Java
// Java program to implement// above approachclass GFG{// Function to check that a number// is the sum of cubes of 2// consecutive numbers or notstatic boolean isSumCube(int N){ int a = (int)Math.cbrt(N); int b = a - 1; // Condition to check if a // number is the sum of cubes of 2 // consecutive numbers or not return ((a * a * a + b * b * b) == N);}// Driver Codepublic static void main(String[] args){ int i = 35; // Function call if (isSumCube(i)) { System.out.print("Yes"); } else { System.out.print("No"); }}}// This code is contributed by Amit Katiyar |
Python3
# Python3 program to # implement above approach # Function to check that a number # is the sum of cubes of 2# consecutive numbers or notdef isSumCube(N): a = int(pow(N, 1 / 3)) b = a - 1 # Condition to check if a # number is the sum of cubes of 2 # consecutive numbers or not ans = ((a * a * a + b * b * b) == N) return ans# Driver Codei = 35# Function callif(isSumCube(i)): print("Yes")else: print("No")# This code is contributed by Shivam Singh |
C#
// C# program to implement// above approachusing System;class GFG{// Function to check that a number// is the sum of cubes of 2// consecutive numbers or notstatic bool isSumCube(int N){ int a = (int)Math.Pow(N, (double) 1 / 3); int b = a - 1; // Condition to check if a // number is the sum of cubes of 2 // consecutive numbers or not return ((a * a * a + b * b * b) == N);}// Driver Codepublic static void Main(String[] args){ int i = 35; // Function call if (isSumCube(i)) { Console.Write("Yes"); } else { Console.Write("No"); }}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to implement// above approach // Function to check that a number// is the sum of cubes of 2// consecutive numbers or notfunction isSumCube(N) { var a = parseInt(Math.cbrt(N)); var b = a - 1; // Condition to check if a // number is the sum of cubes of 2 // consecutive numbers or not return ((a * a * a + b * b * b) == N);}// Driver Codevar i = 35;// Function callif (isSumCube(i)){ document.write("Yes");} else{ document.write("No");}// This code is contributed by todaysgaurav </script> |
Output:
Yes
Time Complexity: O(logN) because using cbrt function
Auxiliary Space: O(1)
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