Integer square root
Given a non-negative integer n (0 <= n <= 2^63 - 1), return floor(sqrt(n)) as an integer, without using any library square-root or float-based math (no math.sqrt, no n**0.5). Floating point loses precision at large n, so use an integer method such as Newton's iteration.
Implement
int_sqrt(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 20 min
solution.py
InputExpectedGot
[8]2not run yetsample