Java Program to Find the Cube Root of a Given Number Using Binary Search

Given a non-negative number find the cube root of a number using the binary search approach.

Examples :

Input: x = 27
Output: 3
Explanation:  The cube root of 16 is 4.

Input: x = 120
Output: 4
Explanation:  The cube root of 120 lies in between
4 and 5 so floor of the cube root is 4.

Naive Approach:

• Check the cube of every element till n and store the answer till the cube is smaller or equal to the n

Java

// Java Program to Find the cube root
// of given number using Naive approach
import java.io.*;
class GFG {
    static int cuberoot(int n)
    {
        int ans = 0;

        for (int i = 1; i <= n; ++i) {
            // checking every number cube
            if (i * i * i <= n) {
                ans = i;
            }
        }
        return ans;
    }
    public static void main(String[] args)
    {
        // Number
        int number = 27;
        // Checking number
        int cuberoot = cuberoot(number);
        System.out.println(cuberoot);
    }
}

Output

3

Complexity:

SpaceComplexity: O(1)
TimeComplexity: O(n)

Efficient Approach (Binary Search): 

Binary Search used Divide and Conquer approach that makes the complexity is O(log n).

Algorithm:

• Initialize left=0 and right =n
• Calculate mid=left+(right-left)/2
• If mid*mid*mid is equal to the number  return the mid
• If mid*mid*mid is less than the number store the mid in ans and increase left=mid+1
• If mid*mid*mid is more than the number and decrease the right=mid-1
• Return the answer

Implementation:

Java

// Java Program to Find the cube root
// of given number using Binary Search
import java.io.*;
import java.util.*;
class GFG {
    // Function to find cuberoot
    static int cuberoot(int number)
    {
        // Lower bound
        int left = 1;
        // Upper bound
        int right = number;

        int ans = 0;
        while (left <= right) {
            // Finding the mid value

            int mid = left + (right - left) / 2;
            // Checking the mid value
            if (mid * mid * mid == number) {
                return mid;
            }

            // Shift the lower bound
            if (mid * mid * mid < number) {
                left = mid + 1;
                ans = mid;
            }
            // Shift the upper bound
            else {
                right = mid - 1;
            }
        }
        // Return the ans
        return ans;
    }
    public static void main(String[] args)
    {
        int number = 215;
        System.out.println(cuberoot(number));
    }
}

Output

5

Complexity:

SpaceComplexity: O(1)
TimeComplexity: O(log n)