Question is
and my solution is
What I think is
Why there is TLE.
Edit 1. I know this is long approach but it is what that I think… Please Help
\bullet line 17: should be
s != 0
instead of
k != 0
[as k is not the one decreasing]
\bullet line 19: should be
if((((k&s) >> --count) & 1)==0)
instead of
if(k&s==0)
[as you are trying to isolate and find whether 1 particular bit is set / unset]
Here I’ve corrected the code and it works just fine and dandy. 
Thank you so much …
You’re welcome 
But what is need of line 19 removing extra things that is --count & 1 code works fine
Let me explain this to you using an example. I’ll be considering nybbles of data.
Let us assume k = 1011 and s = 1111 in binary. Notice the bit position 2:
bit positions:
3|2|1|0
1|0|1|1
1|1|1|1
\&
\rule{1.5cm}{0.4pt}
1|0|1|1 is != 0
Your code stopped checking here.
But, I added the following to isolate bit position 2:
1|0|1|1 >> 2 = 0|0|1|0
and,
0|0|1|0
0|0|0|1
\&
\rule{1.5cm}{0.4pt}
0|0|0|0 is = 0
That’s why we need to shift and then logically and it with 1 for isolating a single bit. Do tell me if you still have anymore doubts.
When k is 1011
Then s is why 1111 it is 1000 because I shift s <<(count-1)
I assumed two arbitrary numbers for k and s in order to explain the bit isolation procedure.
Bur s is not arbitrary in my solution I createD s by shifting left
Then, you may consider s = 0100 to be a valid intermediate value though it starts off with s = 1000, which I didn’t cover as it would give no interesting result and both your and my code would work in that case.
Yes when s become 0100 after shifting loops breaks because I find answer
Corrrrrect!!! 
Thank you so much for giving your precious time
No problem buddy.