Find cubic root of a number

Given a number n, find the cube root of n.

Examples:

Input:  n = 3
Output: Cubic Root is 1.442250

Input: n = 8
Output: Cubic Root is 2.000000

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We can use binary search. First we define error e. Let us say 0.0000001 in our case. The main steps of our algorithm for calculating the cubic root of a number n are:

Initialize start = 0 and end = n
Calculate mid = (start + end)/2
Check if the absolute value of (n – mid*mid*mid) < e. If this condition holds true then mid is our answer so return mid.
If (mid*mid*mid)>n then set end=mid
If (mid*mid*mid)<n set start=mid.

Below is the implementation of above idea.

C++

// C++ program to find cubic root of a number 
// using Binary Search 
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns the absolute value of n-mid*mid*mid 
double diff(double n,double mid) 
{ 
    if (n > (mid*mid*mid)) 
        return (n-(mid*mid*mid)); 
    else
        return ((mid*mid*mid) - n); 
} 
  
// Returns cube root of a no n 
double cubicRoot(double n) 
{ 
    // Set start and end for binary search 
    double start = 0, end = n; 
  
    // Set precision 
    double e = 0.0000001; 
  
    while (true) 
    { 
        double mid = (start + end)/2; 
        double error = diff(n, mid); 
  
        // If error is less than e then mid is 
        // our answer so return mid 
        if (error <= e) 
            return mid; 
  
        // If mid*mid*mid is greater than n set 
        // end = mid 
        if ((mid*mid*mid) > n) 
            end = mid; 
  
        // If mid*mid*mid is less than n set 
        // start = mid 
        else
            start = mid; 
    } 
} 
  
// Driver code 
int main() 
{ 
    double n = 3; 
    printf("Cubic root of %lf is %lf\n", 
           n, cubicRoot(n)); 
    return 0; 
} 

Java

// Java program to find cubic root of a number 
// using Binary Search 
import java.io.*; 
  
class GFG  
{ 
    // Returns the absolute value of n-mid*mid*mid 
    static double diff(double n,double mid) 
    { 
        if (n > (mid*mid*mid)) 
            return (n-(mid*mid*mid)); 
        else
            return ((mid*mid*mid) - n); 
    } 
      
    // Returns cube root of a no n 
    static double cubicRoot(double n) 
    { 
        // Set start and end for binary search 
        double start = 0, end = n; 
   
        // Set precision 
        double e = 0.0000001; 
   
        while (true) 
        { 
            double mid = (start + end)/2; 
            double error = diff(n, mid); 
   
            // If error is less than e then mid is 
            // our answer so return mid 
            if (error <= e) 
                return mid; 
   
            // If mid*mid*mid is greater than n set 
            // end = mid 
            if ((mid*mid*mid) > n) 
                end = mid; 
   
            // If mid*mid*mid is less than n set 
            // start = mid 
            else
                start = mid; 
        } 
    } 
      
    // Driver program to test above function 
    public static void main (String[] args)  
    { 
        double n = 3; 
        System.out.println("Cube root of "+n+" is "+cubicRoot(n)); 
    } 
} 
  
// This code is contributed by Pramod Kumar 

Python3

# Python 3 program to find cubic root  
# of a number using Binary Search 
  
# Returns the absolute value of  
# n-mid*mid*mid 
def diff(n, mid) : 
    if (n > (mid * mid * mid)) : 
        return (n - (mid * mid * mid)) 
    else : 
        return ((mid * mid * mid) - n) 
          
# Returns cube root of a no n 
def cubicRoot(n) : 
      
    # Set start and end for binary  
    # search 
    start = 0
    end = n 
      
    # Set precision 
    e = 0.0000001
    while (True) : 
          
        mid = (start + end) / 2
        error = diff(n, mid) 
  
        # If error is less than e  
        # then mid is our answer 
        # so return mid 
        if (error <= e) : 
            return mid 
              
        # If mid*mid*mid is greater 
        # than n set end = mid 
        if ((mid * mid * mid) > n) : 
            end = mid 
              
        # If mid*mid*mid is less  
        # than n set start = mid 
        else : 
            start = mid 
              
# Driver code 
n = 3
print("Cubic root of", n, "is",  
      round(cubicRoot(n),6)) 
  
  
# This code is contributed by Nikita Tiwari. 

C#

// C# program to find cubic root  
// of a number using Binary Search 
using System; 
  
class GFG { 
      
    // Returns the absolute value 
    // of n - mid * mid * mid 
    static double diff(double n, double mid) 
    { 
        if (n > (mid * mid * mid)) 
            return (n-(mid * mid * mid)); 
        else
            return ((mid * mid * mid) - n); 
    } 
      
    // Returns cube root of a no. n 
    static double cubicRoot(double n) 
    { 
          
        // Set start and end for 
        // binary search 
        double start = 0, end = n; 
  
        // Set precision 
        double e = 0.0000001; 
  
        while (true) 
        { 
            double mid = (start + end) / 2; 
            double error = diff(n, mid); 
  
            // If error is less than e then  
            // mid is our answer so return mid 
            if (error <= e) 
                return mid; 
  
            // If mid * mid * mid is greater  
            // than n set end = mid 
            if ((mid * mid * mid) > n) 
                end = mid; 
  
            // If mid*mid*mid is less than  
            // n set start = mid 
            else
                start = mid; 
        } 
    } 
      
    // Driver Code 
    public static void Main ()  
    { 
        double n = 3; 
        Console.Write("Cube root of "+ n  
                       + " is "+cubicRoot(n)); 
    } 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// PHP program to find cubic root  
// of a number using Binary Search 
  
// Returns the absolute value  
// of n - mid * mid * mid 
function diff($n,$mid) 
{ 
    if ($n > ($mid * $mid * $mid)) 
        return ($n - ($mid *  
                $mid * $mid)); 
    else
        return (($mid * $mid *  
                 $mid) - $n); 
} 
  
// Returns cube root of a no n 
function cubicRoot($n) 
{ 
      
    // Set start and end  
    // for binary search 
    $start = 0;  
    $end = $n; 
  
    // Set precision 
    $e = 0.0000001; 
  
    while (true) 
    { 
        $mid = (($start + $end)/2); 
        $error = diff($n, $mid); 
  
        // If error is less  
        // than e then mid is 
        // our answer so return mid 
        if ($error <= $e) 
            return $mid; 
  
        // If mid*mid*mid is 
        // greater than n set 
        // end = mid 
        if (($mid * $mid * $mid) > $n) 
            $end = $mid; 
  
        // If mid*mid*mid is 
        // less than n set 
        // start = mid 
        else
            $start = $mid; 
    } 
} 
  
    // Driver Code 
    $n = 3; 
    echo("Cubic root of $n is "); 
    echo(cubicRoot($n)); 
  
// This code is contributed by nitin mittal. 
?> 

Output:

Cubic root of 3.000000 is 1.442250

Time Complexity : O(Log n)

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