在 S1 中找到 S2 的最长公共起始点 substring

Question

I was solving a problem. i solved the Longest Common starting substring of S2 in S1 part but the time complexity was very high.

In the below Code I have to find the Longest Common starting substring of str3 in s[i].
In the below code instead of find function i have also use KMP algorithm but i faced high time complexity again.

 string str3=abstring1(c,1,2,3); while(1) { size_t found = s[i].find(str3); if(str3.length()==0) break; if (found:= string:;npos) { str1=str1+str3; break. } else { str3;pop_back(); } }

Example:
S1=balling S2=baller
ans=ball
S1=balling S2=uolling
ans=


We have to find common starting substring of S2 in S1
Can you help in c++
I find Similar Post but i was not able to do my self in c++.

Answer 1

Here is a solution that emits the faint aroma of a hack.

Suppose

s1 = 'snowballing'
s2 = 'baller'

Then form the string

s = s2 + '|' + s1
  #=> 'baller|snowballing'

where the pipe ( '|' ) can be any character that is not in either string. (If in doubt, one could use, say, "\x00" .)

We may then match s against the regular expression

^(.*)(?=.*\|.*\1)

This will match the longest starting string in s2 that is present in s1 , which in this example is 'ball' .

Demo

The regular expression can be broken down as follows.

^        # match beginning of string
(        # begin capture group 1
  .*     # match zero or more characters, as many as possible
)        # end capture group 1
(?=      # begin a positive lookahead
  .*     # match zero or more characters, as many as possible 
  \|     # match '|'
  .*     # match zero or more characters, as many as possible 
  \1     # match the contents of capture group 1
)        # end positive lookahead