Interesting opinions About Job Interviews

I one day stumbled upon this site (Things I Learned From Failing Technical Interviews), a section called "Do not be Yourself" intrigued me.

Contrary to the popular suggestions from many career advisers to be yourself during job interview, the person suggests the interviewee not to be him/herself, but to be a robot.  I totally agree with his opinion.  If you've done multiple job interviews for software engineer job position, you might have experienced some unease when the interviewer expects you to be perfect in coding in front of him/her. No syntax errors, have to memorize the APIs, etc.  The blogger continues to say that hiring a candidate is just like another upgrade of server or adding another PC in the server cluster, which is to offload work from the team to the new hired.

It is almost a guarantee that you will be asked something you don't know and even an hour of thinking about it and problem solving around it you still won't know.  Sometimes you get asked something you've known it long time ago, but can no longer able to recall, grasp it and/or explain it to them clearly.  You feel like you're an idiot with glamorous titles and said experiences (and preconceive that the interviewer must say in his mind, "why the heck this guy is applying for this position", or "Oh boy, another fake"?)

Another job-seeking-related blog: I will get That Job at Google

Swamping two variables without temporary variable

Based on the Boolean Algebra, where XOR operator is commutative and associative:

A = A ⊕ B

B = B ⊕ A = B ⊕ (A ⊕ B) = B ⊕ (B ⊕ A) = (B ⊕ B) ⊕ A

  = 0 ⊕ A = A

A = A ⊕ B = (A ⊕ B) ⊕ A = B ⊕ (A ⊕ A) = B

So the steps are:

A = A^B;

B = B^A;

A = A^B;

Another way is by doing arithmetics:

A = A - B;

B = A + B;

A = B - A;

Or

A = A + B;

B = A - B;

A = A - B;

If we look at carefully, the XOR operator is actually a half-adder operation (with carry bit).

A ⊕ B = A' * B + A * B', where if A=1 the result would be equal to B', else equal to B.

Bits Reversal - Advanced

Suppose we have 16-bit number where we want to reverse some bits between bits i'th and j'th.  To illustrate that, say we have a 16-bit value stored in v:

|<-------- L = sizeof(short)*CHAR_BIT --------->|
 15                       7                1   0
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
|  |  |  | x| x| x| x| x| x| x|  |  |  |  |  |  |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 
          ^                  ^
          |                  |
         i=12               j=6
|<------>|
 L - i - 1
  

So, we want to reverse bits between bits 6th to 12th.

Assume i is always greater than j (i > j):
Number of bits to reverse: n = (i - j) + 1

Mask to use: (2^n - 1) << j

Or, in the case above:

n = (12 - 6) + 1 = 7
mask = (2^7 - 1) << 6 = 127 << 6  = 0x1FC0 (0001111111000000 in binary)

Then we mask the bits stored in v and shift it back to 0 (to do reverse operation):

r = (v & mask)  >> j;

r =
 15                       7                1   0
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
| 0| 0| 0| 0| 0| 0| 0| 0| 0| x| x| x| x| x| x| x|
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+


We use bit reversal operation as posted previously on r:

r = bit_reversal(r);

After this call, the bits are reversed, but filled in from MSB.  We need to put them to the position originally started.

 15                       7                1   0
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
| x| x| x| x| x| x| x|  |  |  |  |  |  |  |  |  |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
                    |-------> shr 3

<========= sizeof(unsigned int)* CHAR_BIT ======>

m = sizeof(short) * CHAR_BIT - n - j
m = 2*8 - 7 -6 = 16 - 13 = 3

Final result is:

(v & !mask) | ((r >> m) & mask);