Re.start() & Re.end() in Python – HackerRank Solution

Re.findall() & Re.finditer() in Python - HackerRank Solution
Re.start() & Re.end() in Python – HackerRank Solution


start() & end()
These expressions return the indices of the start and end of the substring matched by the group.
Code :

>>> import re
>>> m = re.search(r'\d+','1234')
>>> m.end()
4
>>> m.start()
0

Task :

You are given a string S.
Your task is to find the indices of the start and end of string k in S.


Input Format :

The first line contains the string S.
The second line contains the string k.

Constraints :

  • 0 < len(s) < 100
  • 0 < len(k) < len(s)

Output Format :

Print the tuple in this format: (start _index, end _index).
If no match is found, print (-1, -1).


Sample Input :

aaadaa
aa

Sample Output :

(0, 1)
(1, 2)
(4, 5)

Re.start() & Re.end() in Python – HackerRank Solution

import re
S, k = input(), input()
matches = re.finditer(r'(?=(' + k + '))', S)
anymatch = False
for match in matches:
    anymatch = True
    print((match.start(1), match.end(1) - 1))
if anymatch == False:
    print((-1, -1))

Disclaimer: The above Problem (Re.start() & Re.end() in Python – HackerRank Solution) is generated by Hackerrank but the Solution is Provided by Chase2Learn. This tutorial is only for Educational and Learning purposes. Authority if any of the queries regarding this post or website fill the following contact form thank you.

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