Math.sqrt giving floating values in python

Question

I am solving this question:

Given a positive integer num, write a function that returns True if num is a perfect square else False.

Input: num = 16

Output: true

My solution:

import math
class Solution:
    def isPerfectSquare(self, num: int) -> bool:
    
         return type(math.sqrt(16))=='int' //4.0=='int'

I know there are a lot of different solutions which are easier but I want to know how we can get this right as I can't use int to make it integer as 4.5 will also be the correct answer.

Your "solution" appears to be written _only_ for 16. It might as well just `return True`. You're completely ignoring `num`, which should be a hint that you're missing something. — jonrsharpe, Aug 25 '21 at 09:33
@jonrsharpe I'd assume that's just for demonstration/testing/whatever and they'd actually use the `num` argument in the real code. Wouldn't be the first time I've seen something be hardcoded like that just to check something weird wasn't happening with the input value. — Kemp, Aug 25 '21 at 09:34
Or maybe `s = int(round(math.sqrt(num)))` and `return num == s*s` — JohanC, Aug 25 '21 at 09:35
What about `not bool(16**0.5 % 1)`? No libraries and only 1 function call. — S3DEV, Aug 25 '21 at 09:35
https://stackoverflow.com/questions/2489435/check-if-a-number-is-a-perfect-square — umar, Aug 25 '21 at 09:35
@Kemp based on the rest of it that doesn't seem like a safe assumption. — jonrsharpe, Aug 25 '21 at 09:36
@Wolf `class Solution: ...` is the boilerplate used by some online grading platforms. — AKX, Aug 25 '21 at 09:36
It is worth noting that in python>=3.8, there is `math.isqrt` which computes the floor of the square root of any nonnegative integer (including integers too large for floating-point calculations) — Stef, Aug 25 '21 at 09:58
Python code for `isqrt`, integer square root: https://github.com/python/cpython/blob/main/Modules/mathmodule.c#L1562 — Stef, Aug 25 '21 at 10:01

score -1 · Answer 1 · edited Aug 25 '21 at 10:49

-1

math.sqrt's default return type is float so you should check if the square root and floor of square root are equal or not to identify a perfect square...

This will work...

import math
class Solution:
    def isPerfectSquare(self, num: int) -> bool:
        sqroot = math.sqrt(num)
        return sqroot == math.floor(sqroot)

edited Aug 25 '21 at 10:49

Sapt-Programmer

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answered Aug 25 '21 at 09:38

MohammadArik

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You shouldn't be using `==` with floats. See [Is floating point math broken](https://stackoverflow.com/questions/588004/is-floating-point-math-broken). (Also, using `sqrt` as a variable name isn't a good idea.) – JohanC Aug 25 '21 at 09:45
1

Better use `round(sqroot) ** 2 == num` and tell about the restriction of `num` to something around `2**26` – Wolf Aug 25 '21 at 11:38
@Wolf thats also doable but the current solution will is faster. It wont matter in small numbers but in time of big numbers, it will be a problem.. – MohammadArik Aug 26 '21 at 01:54

Math.sqrt giving floating values in python

1 Answers1