Application Engineer Interview Questions

33%

4/9

From a 3x3x3 Cube you can make 27 1x1x1 cubes, Since one face of the cube is painted blue. once face of 9 1x1 cubes will be blue. hence the answer is 9/27 = 1/3 = 33%

10=2 100=4 1000=8… Caught the thing .. he didn't like my implementation. don't be nervous!

return (( x!=0) && (x & (x -1) == 0)) a number which is power of two will have single 1 in binary representation ex. 16 00010000 x-1 will have pattern 00001111 hence x&(x-1) will always be 0 for power of two numbers. Need to check condition for x!=0 since 0 is not a power of two. Special case to handle.

if you can figure this out inside an hour, congratulations this isnt something that should be in an interview to be honest the level of difficulty of the answer that algo came up with is at the level of an upper division course that has a proof of it written in wikipedia i honestly like the brute force solution

because the lids for man-holes are round...

Because they can be easily rolled over. BTW, have you completed you technical interview. Please share the questions. Thanks.

So the lid wont drop down into the hole! With rectangular man holes, there is always a way to do so

Impedance to provide for maximum transmission of power.

It is trying to match both the real and reactive source or load impedance to the system Zo, usually. Doing so provides maximum gain, and in a purely linear system maximum power transfer Matching networks may also be quite commonly used to match the system Zo (say 50 Ohms) to the load line of a non-linear device such as a power amplifier, or just as similarly to match tot he optimum noise impedance of an LNA. In these cases a perfect match is not desired, as maximum gain is not the goal, but rather maximum efficiency or lowest noise figure.

You want the real parts to match and the reactive components to cancel if you want max power transfer

which website are you referring to?

Glassdoor

could you please post the link for those questions? I'm new to this website and i have not figured out how to find the questions.

x=rand(1000000,1); y=10*x; x=ceil(y); y=length(x); a=1; for z=1:y for z1=1:y if (x(z)+x(z1))== 10 r(a,1)=x(z); r(a,2)=x(z1); a=a+1; end end end MATLAB'd

& if they want non repetitive numbers: [code cont'd] g=size(r); b=g(1,1)-1; c=g(1,2); q=1; for d=1:b if r(d,1)~=r(d+1,1) && r(d,2)~=r(d+1,1) l(q,1)=r(d,1); l(q,2)=r(d,2); q=q+1; end end

Not really optimal. I'll give you my approach: 1) First you do binary search, to located the Last 10. 2) Create 2 pointers, 1 at the start, and the other from the last 10. 3) Bring them closer together creating a new array of all pairs summing up to 10. 4) If 1 pointer gets out of sync, save the index of that pointer (only once), and skip all same numbers till the pointers get in sync again. 5) Recurse within all the indices you skipped. 6) Return the list of pairs.

80 hours.

Let us suppose that we have 2 different clocks with the faulty one completing 51 mins in an hour. -the two clocks will again show the same time when the lag time = 12 hours for 1 iteration=>lag time =9 mins for getting a lag time of 12 hours no of iterations needed = 12*60/9=80 one iteration means 1 hour so the next time the clocks will be in sync= 80 hours time at that point=8:00 PM

if it is analog it takes 40 hours (6 hours late, and 6 hours passed) if it is digital 80 hours.