I had the following question in an interview and I believe I gave a working implementation but I was wondering if there was a better implementation that was quicker, or just a trick I missed.
Given 3 unsigned 30-bit integers, return the number of 30-bit integers that when compared with any of the original numbers have the same position bits set to 1. That is we enumerate all the 0s
Let me give an example, but lets use 4bit for clarity.
A = 1001 B = 0011 C = 0110
It should return 8 since there are 8 4bit ints that are in the set. The set being:
0011 0110 0111 1001 1011 1101 1110 1111
Now how I worked it out was to take each number and enumerate the set of possibilites and then count all the distinct values. How I enumerated the set was to start with the number, add one to it and then OR it with itself until I reached the mask. With the number itself being in the set and the mask (being all set to 1) also in the set. So for example to enumerate the set of 1001:
1001 = the start 1011 = (1001 + 1) | 1001 1101 = (1011 + 1) | 1001 1111 = (1101 + 1) | 1001 (this is the last value as we have reached our mask)
So do that for each number and then count the uniques.
This is it in python code (but language doesn’t really matter as long as you can do bitwise operations, hence why this question is tagged for c/c++ as well):
MASK = 0x3FFFFFFF def count_anded_bitmasks( A, B, C ): andSets = set( enumerate_all_subsets(A) + enumerate_all_subsets(B) + enumerate_all_subsets(C) ) return len(andSets) def enumerate_all_subsets( d ): andSet =  n = d while n != MASK: andSet.append(n) n = (n + 1) | d andSet.append(n) return andSet
Now this works and gives the correct answer but I’m wondering if I have missed a trick. Since the question was to only ask the count and not enumerate all the values perhaps there is a much quicker way. Either by combining the numbers first, or getting a count without enumeration. I have a feeling there is. Since numbers that contain lots of zeros the enumeration rises exponentially and it can take quite a while.
If you have A B and C, the count of the set of numbers which has bits set to 1 where A or B or C has corresponding bits set to 1.
Some people don’t understand the question (didn’t help that I didn’t ask it correctly first of). Let’s use the given A B and C values above:
1001 1011 1101 1111
0011 0111 1011 1111
0110 0111 1110 1111
Now combine those sets and count the distinct entries. That is the answer. Is there a way to do this without enumerating the values?
edit: Sorry for the mistake the question. Fixed now.
N = 4 def supers(number): zeros = sum(1 for bit in xrange(N) if (number >> bit) & 1 == 0) return 2**zeros def solve(a,b,c): total = supers(a) + supers(b) + supers(c) total -= supers(a | b) # counted twice, remove one total -= supers(b | c) # counted twice, remove one total -= supers(a | c) # counted twice, remove one total += supers(a | b | c) # counted three times, removed three times, add one return total print solve(0b1001,0b0011,0b0110)
S(n) be the set produce by the number
|S(n)| the size of the set for the number n.
supers is not a great name, but I had trouble coming up with a better one
The trick is to realize that
S(a) ^ S(b) = S(a | b). As a result, using supers I can figure out the size of all those sets.
To figure out the rest, draw a venn diagram of the sets.