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  1. Prepare
  2. Python
  3. Regex and Parsing
  4. Re.start() & Re.end()
  5. Discussions

Re.start() & Re.end()

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349 Discussions

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  • 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

  • akshaynambiar08
     + 0 comments

    You know what, I don't care about re on this one. It broke my brain. But here is a version that passed all the test cases without re: if name == "main": s: str = input() k: str = input()

    found = False
for i in range(len(s)):
    if k == s[i: i + len(k)]:
        print((i, i + len(k) - 1))
        found = True
if not found:
    print((-1, -1))

  • reddysantosh1310
     + 0 comments
    # Enter your code here. Read input from STDIN. Print output to STDOUT

import re

s = input()
k = input() 
regex_pattern =  r'(?={})'.format(re.escape(k))
found = False
for match in re.finditer(regex_pattern,s):
    
    print((match.start(),match.start() + len(k) - 1))
    found = True
    
if not match:
    print((-1,-1))