Sqrt(x)

Problem Statement:

Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well. You must not use any built-in exponent function or operator.

• For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.

Example 1:

Input: x = 4 Output: 2 Explanation: The square root of 4 is 2, so we return 2.

Example 2:

Input: x = 8 Output: 2 Explanation: The square root of 8 is 2.82842…, and since we round it down to the nearest integer, 2 is returned.

Constraints: 🌟🌟

• 0 $<=$ x $<=$ 2^3$$^1 - $1$

Solution:

Easy one. First we need to search for minimal k satisfying condition k^2 > x, then k - 1 is the answer to the question. We can easily come up with the solution. Notice that I set right = x + 1 instead of right = x to deal with special input cases like x = 0 and x = 1.

class Solution {
	public boolean condition(long square, long val){
		if(val * val > square){
			return true;
		} else{
			return false;
		}
	}
	public int mySqrt(int x) {
		long start = 0;
		long end = (long)x + 1;
	
		while(start < end){
			long mid = start + (end - start)/2;
	
			if(condition(x, mid)){
				end = mid;
			} else{
				start = mid + 1;
			}
		}
		return (int)start - 1; //`start` is the minimum k value s.t. k^2 > x, Therefor `k - 1` is the answer
	}
}