Number of times the largest Perfect Cube can be subtracted from N

Given a number N, at every step, subtract the largest perfect cube( ≤ N) from N. Repeat this step while N > 0. The task is to count the number of steps that can be performed.

Examples:

Input: N = 100
Output: 4
First step, 100 – (4 * 4 * 4) = 100 – 64 = 36
Second step, 36 – (3 * 3 * 3) = 36 – 27 = 9
Third step, 9 – (2 * 2 * 2) = 9 – 8 = 1
Fourth step, 1 – (1 * 1 * 1) = 1 – 1 = 0

Input: N = 150
Output: 5
First step, 150 – (5 * 5 * 5) = 150 – 125 = 25
Second step, 25 – (2 * 2 * 2) = 25 – 8 = 17
Third step, 17 – (2 * 2 * 2) = 17 – 8 = 9
Fourth step, 9 – (2 * 2 * 2) = 9 – 8 = 1
Fifth step, 1 – (1 * 1 * 1) = 1 – 1 = 0

Approach:



  1. Get the number from which the largest perfect cube has to be reduced.
  2. Find the cube root of the number and convert the result as an integer. The cube root of the number might contain some fraction part after the decimal, which needs to be avoided.
  3. Subtract the cube of the integer found in the previous step. This would remove the largest possible perfect cube from the number in the above step.
    N = N - ((int) ∛N)3
    
  4. Repeat the above two steps with the reduced number, till it is greater than 0.
  5. Print the number of times a perfect cube has been reduced from N. This is the final result.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of steps
int countSteps(int n)
{
  
    // Variable to store the count of steps
    int steps = 0;
  
    // Iterate while N > 0
    while (n) {
  
        // Get the largest perfect cube
        // and subtract it from N
        int largest = cbrt(n);
        n -= (largest * largest * largest);
  
        // Increment steps
        steps++;
    }
  
    // Return the required count
    return steps;
}
  
// Driver code
int main()
{
    int n = 150;
    cout << countSteps(n);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG{
   
// Function to return the count of steps
static int countSteps(int n)
{
   
    // Variable to store the count of steps
    int steps = 0;
   
    // Iterate while N > 0
    while (n > 0) {
   
        // Get the largest perfect cube
        // and subtract it from N
        int largest = (int) Math.cbrt(n);
        n -= (largest * largest * largest);
   
        // Increment steps
        steps++;
    }
   
    // Return the required count
    return steps;
}
   
// Driver code
public static void main(String[] args)
{
    int n = 150;
    System.out.print(countSteps(n)); 
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 implementation of the approach
from math import floor
  
# Function to return the count of steps
def countSteps(n):
  
    # Variable to store the count of steps
    steps = 0
  
    # Iterate while N > 0
    while (n):
  
        # Get the largest perfect cube
        # and subtract it from N
        largest = floor(n**(1/3))
        n -= (largest * largest * largest)
  
        # Increment steps
        steps += 1
  
    # Return the required count
    return steps
  
# Driver code
n = 150
print(countSteps(n))
  
# This code is contributed by mohit kumar 29

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C#

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// C# implementation of the approach
using System;
  
class GFG{
    
// Function to return the count of steps
static int countSteps(int n)
{
    
    // Variable to store the count of steps
    int steps = 0;
    
    // Iterate while N > 0
    while (n > 0) {
    
        // Get the largest perfect cube
        // and subtract it from N
        int largest = (int) Math.Pow(n,(double)1/3);
        n -= (largest * largest * largest);
    
        // Increment steps
        steps++;
    }
    
    // Return the required count
    return steps;
}
    
// Driver code
public static void Main(String[] args)
{
    int n = 150;
    Console.Write(countSteps(n)); 
}
}
  
// This code is contributed by PrinciRaj1992

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Output:

5

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