Sqrt(x)
Signals to notice
Brute force first
Try every integer from 1 upward until i² > x. Linear search through candidates. That direct path helps you understand the question, but it tends to treat every possibility as brand new instead of learning from earlier steps.
The key insight
Binary search on the range [1, x]: if mid² ≤ x, the answer is mid or higher; if mid² > x, it's lower. Converges in steps. Once you hold onto the right piece of information from moment to moment, the problem feels less like trial and error and more like following a shape that was there all along.
What must stay true
The answer is the largest integer m where m² ≤ x. Binary search narrows the range by half each step, and the left boundary always stays ≤ the answer. When you keep that truth intact, each local choice supports the larger solution instead of fighting it.
Easy way to go wrong
Integer overflow when computing mid * mid — use long/BigInt or compare mid vs x/mid instead of mid*mid vs x. Most mistakes here are not about syntax; they come from losing track of what your state, pointer, or structure is supposed to mean.