Integer square root
Given a non-negative integer n, return the floor of its square root (the largest integer r with r*r <= n) without using any floating-point square-root function. Handle n = 0 and n = 1.
Implement
isqrt_floor(n: int) → intExamples
in
[8]out2What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 15 min
solution.py
InputExpectedGot
[8]2not run yetsample