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 = 0Input: 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:
- Get the number from which the largest perfect cube has to be reduced.
- 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.
- 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
- Repeat the above two steps with the reduced number, till it is greater than 0.
- 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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count of stepsint 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 codeint main(){ int n = 150; cout << countSteps(n); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function to return the count of stepsstatic 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 codepublic static void main(String[] args){ int n = 150; System.out.print(countSteps(n)); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approachfrom math import floor# Function to return the count of stepsdef 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 coden = 150print(countSteps(n))# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the count of stepsstatic 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 codepublic static void Main(String[] args){ int n = 150; Console.Write(countSteps(n)); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript implementation of the approach// Function to return the count of stepsfunction countSteps(n){ // Variable to store the count of steps let steps = 0; // Iterate while N > 0 while (n) { // Get the largest perfect cube // and subtract it from N let largest = Math.floor(Math.cbrt(n)); n -= (largest * largest * largest); // Increment steps steps++; } // Return the required count return steps;}// Driver codelet n = 150;document.write(countSteps(n));// This code is contributed by Manoj.</script> |
Output:
5
Time complexity: O(logn), as using inbuilt cbrt function
Auxiliary space: O(1)
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