What is quickest way to determine if a given binary number is a perfect square??

Question

Input is given as string(length N) in binary form(0 & 1)

Output will be in form of Yes or No
0<N<100

Ex. 1)
Input - 100
Output - Yes

Input - 1000000111
Output - No

Answer 1

● basic way is to convert it from binary to decimal number a & then apply this condition (int)(Math.sqrt(a)*Math.sqrt(a))==a).

//I found a fast way to identify if a number is not a square(it will not confirm square number, but it will confirm if a number is not square number)
● if any of 1) or 2) condition becomes false, then number will not be perfect square else number may or may not be perfect square(you have to check by above basic way).
1)check the last digit, it should end with 1,4,5,6,9, 0 (end with even number of zeroes but lets not increase the code lines & steps in it)
//if 1) comes true, then move for 2)
2)check for the recursive sum of digits (keep doing sum of digits intil a single number comes) if it is 1,4,7,9 then number may be a square number but if it is not, then it is confirmed it will not be a square number.

//so 1) will take very less time in checking for non-square number, after that I have still doubt in 2) & basic way which will be fast & efficient.

Answer 2

In Python it should be:

1 def check(num):
2 print(f"Input - {num}")
3 num = int(num, 2)
4 print(f"Output - {'No' if num**0.5 - int(num**0.5) else 'Yes'}")
5
6 check(input())
7