Help me in solving LUCNUM problem

help
,
1

My issue

In the given test cases , for very first test, 1 as input , the answer is 1 as output , similarly in my code , for t = 1 , n =1 , output is showing 1 , but the judge is showing there is no expected output for this problem as there can be multiple correct outputs.

My code

for i in range(int(input())):
    n = int(input())
    i = 0
    while(n):
        if n%2 == 0 : 
            i= not i 
            n/=2 
        else:
            break
    print(0 if i else 1)

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2

@codingninja_x
your logic is not right .
don’t use debug your code for this problem .
Your code is failing for some else case.
plzz refer the following python code for better understanding

T = int(input())

def lucky(N):
    count = 0 
    while N % 2 == 0:
        N = N//2
        count += 1 
    return 1 - (count % 2)

for t in range(T):
    N = int(input())
    print(lucky(N))
3

@dpcoder_007
After thoroughly comparing the codes ,

The only place our code differs is :
Your code : counts the number of times 2’s power is there and returning the answer by taking 1-count%2.
My Code : flips the bit based on the count of 2’s power in the number N .

Edge case which i can think of only here is 1 which is getting covered in both your code and my code.

I saw and ran the code for input 1 to 10 for both of ours code , still the resultant is same in each case.

All I want is a test case where my logic is getting wrong. It’s fine if it’s against the policy then you can mention so , but still I am not able to figure out what’s the difference in logic apart from the one i mentioned above.