We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Python
  3. Regex and Parsing
  4. Re.start() & Re.end()
  5. Discussions

Re.start() & Re.end()

Sort by

recency

|

351 Discussions

|

  • tusharvyavahare1
     + 0 comments
    import re
s=input()
k=input()
flag=False
for i in re.finditer(rf'(?=({k}))',s):
    #print(i.group(1))
    print((i.start(1),i.end(1)-1))
    flag=True
if not flag:
    print((-1,-1))

  • mohamedhari317
     + 0 comments
    import re

s = input()
k = input()
regex_pattern =  r'(?={})'.format(re.escape(k))


l=list(re.finditer(regex_pattern,s))
if l==[]:
    print((-1,-1))
else:
    for match in l:
        print((match.start(),match.start()+len(k)-1))

  • jakubszrajber12
     + 0 comments
    import re

s = input()
k = re.escape(input())

pattern = f'(?={k})'
lenght = len(k)

matches = re.finditer(pattern, s)
checker = list(matches)

if checker:
    for m in checker:
        start = m.start()
        end = start + lenght - 1
        print((start, end))
else:
    print((-1, -1))

  • amiraazizan23
     + 0 comments

    import re

    S = input() k = input()

    pattern = re.compile(f'(?={re.escape(k)})') matches = list(pattern.finditer(S))

    if matches: for m in matches: print((m.start(), m.start() + len(k) - 1)) else: print((-1, -1))

  • yashdeora98294
     + 0 comments

    Here is HackerRank Re.start() & Re.end() in Python solution - https://programmingoneonone.com/hackerrank-re-start-re-end-solution-in-python.html