Given a non-negative number find the square root of a number using the binary search approach.
Examples :
Input: x = 16 Output: 4 Explanation: The square root of 16 is 4. Input: x = 5 Output: 2 Explanation: The square root of 5 lies in between 2 and 3 so floor of the square root is 2.
Naive Approach:
- Check the square of every element till n and store the answer till the square is smaller or equal to the n
Java
// Java program to Find the square root of given numbers// by brute force techniqueÂ
import java.io.*;Â
class GFG {Â Â Â Â static int cuberoot(int n)Â Â Â Â {Â Â Â Â Â Â Â Â int ans = 0;Â
        for (int i = 1; i <= n; ++i) {                       // checking every number cube            if (i * i <= n) {                ans = i;            }        }        return ans;    }    public static void main(String[] args)    {        // Number        int number = 16;               // Checking number        int cuberoot = cuberoot(number);        System.out.println(cuberoot);    }} |
Output
4
SpaceComplexity: O(1)
TimeComplexity: O(n)
Efficient Approach (Binary Search): Binary Search used Divide and Conquer approach that makes the complexity is O(logn).
Algorithm:
- Initialize left=0 and right =n
- Calculate mid=left+(right-left)/2
- If mid*mid is equal to the number return the mid.
- If mid*mid is less than the number store the mid in ans since this can possibly be the answer and increase left=mid+1 and now check in the right half.
- If mid*mid is more than the number and decrease the right=mid-1 since the expected value is lesser therefore we will now look into the left half part or will be scanning the smaller values.
- Return the answer
Implementation:
Java
// Java program to Find the square root of given numbers// using Binary searchÂ
// Importing librariesimport java.io.*;import java.util.*;class GFG {       // Function to find cuberoot    static int squareeroot(int number)    {        // Lower bound        int left = 1;               // Upper bound        int right = number;Â
        int ans = 0;        while (left <= right) {                       // Finding the mid valueÂ
            int mid = left + (right - left) / 2;                       // Checking the mid value            if (mid * mid == number) {                return mid;            }Â
            // Shift the lower bound            if (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 = 15;        System.out.println(squareroot(number));    }} |
Output
3
Time Complexity: O(logn)
Auxiliary space: O(1) as it is using constant variables
