Least number to be added to or subtracted from N to make it a Perfect Cube

Namachila 1 hour ago

0 0 4 minutes read

Given a number

, Find the minimum number that needs to be added to or subtracted from

, to make it a

perfect cube

. If the number is to be added, print it with a + sign, else if the number is to be subtracted, print it with a – sign.

Examples:

Input: N = 25 Output: 2 Nearest perfect cube before 25 = 8 Nearest perfect cube after 25 = 27 Therefore 2 needs to be added to 25 to get the closest perfect cube Input: N = 40 Output: -13 Nearest perfect cube before 40 = 25 Nearest perfect cube after 40 = 64 Therefore 13 needs to be subtracted from 40 to get the closest perfect cube

Approach

Get the number.
Find the cube root of the number and convert the result as an integer.
After converting the double value to integer, this will contain the root of the perfect cube before N, i.e. floor(cube root(N)).
Then find the cube of this number, which will be the perfect cube before N.
Find the root of the perfect cube after N, i.e. the ceil(cube root(N)).
Then find the cube of this number, which will be the perfect cube after N.
Check whether the cube of floor value is nearest to N or the ceil value.
If the cube of floor value is nearest to N, print the difference with a -sign. Else print the difference between the cube of the ceil value and N with a + sign.

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 Least number
int nearest(int n)
{
 
    // Get the perfect cube
    // before and after N
    int prevCube = cbrt(n);
    int nextCube = prevCube + 1;
    prevCube = prevCube * prevCube * prevCube;
    nextCube = nextCube * nextCube * nextCube;
 
    // Check which is nearest to N
    int ans
        = (n - prevCube) < (nextCube - n)
              ? (prevCube - n)
              : (nextCube - n);
 
    // return the result
    return ans;
}
 
// Driver code
int main()
{
    int n = 25;
    cout << nearest(n) << endl;
 
    n = 27;
    cout << nearest(n) << endl;
 
    n = 40;
    cout << nearest(n) << endl;
 
    return 0;
}

Java

// Java implementation of the approach 
class GFG {
     
    // Function to return the Least number 
    static int nearest(int n) 
    { 
     
        // Get the perfect cube 
        // before and after N 
        int prevCube = (int)Math.cbrt(n); 
        int nextCube = prevCube + 1; 
        prevCube = prevCube * prevCube * prevCube; 
        nextCube = nextCube * nextCube * nextCube; 
     
        // Check which is nearest to N 
        int ans = (n - prevCube) < (nextCube - n) ? 
                    (prevCube - n) : (nextCube - n); 
     
        // return the result 
        return ans; 
    } 
     
    // Driver code 
    public static void main (String[] args)
    { 
        int n = 25; 
        System.out.println(nearest(n)); 
     
        n = 27; 
        System.out.println(nearest(n)) ; 
     
        n = 40; 
        System.out.println(nearest(n)) ; 
    } 
}
 
// This code is contributed by Yash_R

Python3

# Python3 implementation of the approach
 
import math
 
# Function to return the least number
def nearest(n):
    # Get the perfect cube before and after N
    prev_cube = int(n ** (1/3))
    next_cube = prev_cube + 1
    prev_cube = prev_cube ** 3
    next_cube = next_cube ** 3
 
    # Check which is nearest to N
    ans = prev_cube - n if n - prev_cube < next_cube - n else next_cube - n
 
    # Return the result
    return ans
 
 
# Driver code
if __name__ == "__main__":
    n = 25
    print(nearest(n))
 
    n = 27
    print(nearest(n))
 
    n = 40
    print(nearest(n))
 
 
# by phasing17

C#

using System;
class GFG {
 
    // Function to return the least number
    static int Nearest(int n)
    {
        // Get the perfect cube before and after N
        int prevCube = (int)(Math.Pow(n, 1.0 / 3.0));
        int nextCube = prevCube + 1;
        prevCube = prevCube * prevCube * prevCube;
        nextCube = nextCube * nextCube * nextCube;
 
        // Check which is nearest to N
        int ans = (n - prevCube) < (nextCube - n)
                      ? (prevCube - n)
                      : (nextCube - n);
 
        // return the result
        return ans;
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        int n = 25;
        Console.WriteLine(Nearest(n));
 
        n = 27;
        Console.WriteLine(Nearest(n));
 
        n = 40;
        Console.WriteLine(Nearest(n));
    }
}

Javascript

// Javascript implementation for the approach
function nearest(n) {
    // Get the perfect cube before and after N
    const prevCube = Math.floor(Math.cbrt(n));
    const nextCube = prevCube + 1;
    const prevCubeValue = prevCube * prevCube * prevCube;
    const nextCubeValue = nextCube * nextCube * nextCube;
 
    // Check which is nearest to N
    const ans = Math.abs(n - prevCubeValue) < Math.abs(nextCubeValue - n)
        ? (prevCubeValue - n)
        : (nextCubeValue - n);
 
    // Return the result
    return ans;
}
 
// Driver code
const n1 = 25;
console.log(nearest(n1));
 
const n2 = 27;
console.log(nearest(n2));
 
const n3 = 40;
console.log(nearest(n3));
 
// This code is contributed by Taranpreet Singh.

Output

2
0
-13

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