LeetCode #69: Sqrt(x) | Binary Search

0:00 Problem overview
0:35 O(sqrt x) solution
1:10 Binary Search O(log n) solution
2:10 Explanation of O(log n) runtime
3:00 Code walkthrough

#leetcode #programming #coding #algorithms

[singhamita]

thanks for sharing intution and approach nicely

[dnm]

Keep it up man! You’re really good with these

[mohamedelmahy]

My man, you're by far the best one explaining Leetcode questions. Please keep up the good work, i struggle when i don't find a question in your channel. Please do all 2k+ :)

[Sai-epon]

Bro your work is so good, pleace concentrate on completing the blind 75 playlist, Which will be more helpful
-Thanks for your work

[vikrant_bhatt]

4:03 it is showing run-time error
for 0

[weiwang]

Is it possible to solve by:

mid = left + right) / 2;
if ((mid * mid <= x) {
left = mid;
} else if (mid * mid >= x) {
right = mid;
left = 0;
}

This yields results like x=9, left = right = 3.00001 but there is no way to decide if the final return is 2 or 3 because it infinitely closes to the actual value but how far left to the exact answer VS right to the exact answer cannot be decided. Is it possible to build a solution based on above logic? Below is the return condition I have:

if (left * left < x && right * right > x && right - left <= 0.01) {
return (int) Math.floor(right);
}

which does not work no matter how small the epsilon value is.