Codility is a popular automated test site of programming skills. By taking its tests you will demonstrate your programming ability on some small coding assignments. Presented with a few tasks, you will require to write or correct some code. Here I want to show some lessons you can find on Codility and their solutions, in Python.
A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.
For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps. The number 32 has binary representation 100000 and has no binary gaps.
Write a function:
def solution(N)
that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn’t contain a binary gap.
For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5. Given N = 32 the function should return 0, because N has binary representation ‘100000’ and thus no binary gaps.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..2,147,483,647].
This function applies some important concepts about regular expressions that need to be sharedin advance: lookahead and lookbehind, collectively called lookarounds.
Lookarounds are Regular Expressions that actually match characters, but then give up the match, returning only the result: match or no match.
It may sound confusing, but basically they don’t “use” the matched character, allowing us to use it again on the next expression check:
import re
def solution(N):
zeros = re.findall(r'(?<=1)0+(?=1)', bin(N)[2:])
return len(max(zeros)) if zeros else 0
This is specially usefull for this case since allow us to match zeroes in between ones without “using” such ones. E.g: for 10001001 we will have two matches, 000 and 00. In order for this to work we have to use the 1 in the middle twice. We get this by applying lookbehind (?<=X) and lookahed (?=X), being X the match we want, in this case 1.
To be continued...