Group(), Groups() & Groupdict()

group()

group() expression returns one or more subgroups of the match.
Code

>>> import re
>>> m = re.match(r'(\w+)@(\w+)\.(\w+)','username@hackerrank.com')
>>> m.group(0)       # The entire match 
'username@hackerrank.com'
>>> m.group(1)       # The first parenthesized subgroup.
'username'
>>> m.group(2)       # The second parenthesized subgroup.
'hackerrank'
>>> m.group(3)       # The third parenthesized subgroup.
'com'
>>> m.group(1,2,3)   # Multiple arguments give us a tuple.
('username', 'hackerrank', 'com')

groups()

groups() expression returns a tuple containing all the subgroups of the match.
Code

>>> import re
>>> m = re.match(r'(\w+)@(\w+)\.(\w+)','username@hackerrank.com')
>>> m.groups()
('username', 'hackerrank', 'com')

groupdict()

groupdict() expression returns a dictionary containing all the named subgroups of the match, keyed by the subgroup name.
Code

>>> m = re.match(r'(?P<user>\w+)@(?P<website>\w+)\.(?P<extension>\w+)','myname@hackerrank.com')
>>> m.groupdict()
{'website': 'hackerrank', 'user': 'myname', 'extension': 'com'}

Task

You are given a string S.
Your task is to find the first occurrence of an alphanumeric character in S (read from left to right) that has consecutive repetitions.

Input Format

A single line of input containing the string S.

Constraints

0 < len(S) < 100

Output Format

Print the first occurrence of the repeating character. If there are no repeating characters, print -1.

Sample Input

..12345678910111213141516171820212223

Sample Output

1

Explanation

.. is the first repeating character, but it is not alphanumeric.
1 is the first (from left to right) alphanumeric repeating character of the string in the substring 111.

Solution Implementation


import re
# Can't use '\w' since you dont want to match on underscore..
m = re.search(r'([a-zA-Z0-9])\1+', input().strip())
print(m.group(1) if m else -1)
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