Check if a number can be represented as sum of two consecutive perfect cubes

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++

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// 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 numbers
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 Code
int main()
{
    int n = 35;
 
    if (isCubeSum(n))
        cout << "Yes";
    else
        cout << "No";
}

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Java

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// Java program of the
// above approach
import java.util.*;
 
class GFG{
 
// Function to check if a number
// can be expressed as the sum of
// cubes of two consecutive numbers
static 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 Code
public 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

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Python3

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# Python3 program of the
# above approach
 
# Function to check if a number
# can be expressed as the sum of
# cubes of two consecutive numbers
def 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 Code
if __name__ == '__main__':
     
    n = 35;
 
    if (isCubeSum(n)):
        print("Yes");
    else:
        print("No");
 
# This code is contributed by Amit Katiyar

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C#

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// C# program of the
// above approach
using System;
 
class GFG{
 
// Function to check if a number
// can be expressed as the sum of
// cubes of two consecutive numbers
static 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 Code
public 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

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Output: 

Yes






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

  • For example, if N = 35, then check of the equation below os equal to N or not:

\lfloor \sqrt[3]{35} - 1 \rfloor ^3 + \lfloor \sqrt[3]{35} \rfloor^3

Below is the implementation of the above approach:

C++

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// 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 not
bool 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 Code
int main()
{
    int i = 35;
    // Function call
    if (isSumCube(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

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Java

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// Java program to implement
// above approach
class GFG{
 
// Function to check that a number
// is the sum of cubes of 2
// consecutive numbers or not
static 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 Code
public 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

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Python3

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# Python3 program to
# implement above approach
 
# Function to check that a number
# is the sum of cubes of 2
# consecutive numbers or not
def 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 Code
i = 35
 
# Function call
if(isSumCube(i)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Shivam Singh

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C#

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// C# program to implement
// above approach
using System;
class GFG{
 
// Function to check that a number
// is the sum of cubes of 2
// consecutive numbers or not
static 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 Code
public 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

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Output: 

Yes






Time Complexity: O(1) 
Auxiliary Space: O(1) 

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