Question

If I have an input string and an array:

s = "to_be_or_not_to_be" 
pos = [15, 2, 8]

I am trying to find the longest common prefix between the consecutive elements of the array pos referencing the original s. I am trying to get the following output:

longest = [3,1]

The way I obtained this is by computing the longest common prefix of the following pairs:

• s[15:] which is _be and s[2:] which is _be_or_not_to_be giving 3 ( _be )
• s[2:] which is _be_or_not_to_be and s[8:] which is _not_to_be giving 1 ( _ )

However, if s is huge, I don’t want to create multiple copies when I do something like s[x:]. After hours of searching, I found the function buffer that maintains only one copy of the input string but I wasn’t sure what is the most efficient way to utilize it here in this context. Any suggestions on how to achieve this?

Answer 1

Here is a method without buffer which doesn’t copy, as it only looks at one character at a time:

from itertools import islice, izip

s = "to_be_or_not_to_be"
pos = [15, 2, 8]


length = len(s)    

for start1, start2 in izip(pos, islice(pos, 1, None)):
    pref = 0
    for pos1, pos2 in izip(xrange(start1, length), xrange(start2, length)):
        if s[pos1] == s[pos2]:
            pref += 1
        else:
            break
    print pref
# prints 3 1

I use islice, izip, and xrange in case you’re talking about potentially very long strings.

I also couldn’t resist this “One Liner” which doesn’t even require any indexing:

[next((i for i, (a, b) in 
    enumerate(izip(islice(s, start1, None), islice(s, start2, None))) 
        if a != b), 
    length - max((start1, start2))) 
 for start1, start2 in izip(pos, islice(pos, 1, None))]

One final method, using os.path.commonprefix:

[len(commonprefix((buffer(s, n), buffer(s, m)))) for n, m in zip(pos, pos[1:])]