My answer, pretty slow, but very easy to understand. Working on a file with 100 strings of 1 kb each takes about two seconds, returns any one longest substring if there are more than one
ls = list()
ls.sort(key=len)
s1 = ls.pop(0)
maxl = len(s1)
#1 create a list of all substrings backwards sorted by length. Thus we don't have to check the whole list.
subs = [s1[i:j] for i in range(maxl) for j in range(maxl,i,-1)]
subs.sort(key=len, reverse=True)
#2 Check a substring with the next shortest then the next etc. if is not in an any next shortest string then break the cycle, it's not common. If it passes all checks, it is the longest one by default, break the cycle.
def isasub(subs, ls):
for sub in subs:
for st in ls:
if sub not in st:
break
else:
return sub
break
print('the longest common substring is: ',isasub(subs,ls))