Table of Contents

Group(), Groups() & Groupdict() hackerrank

_group()

A 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()

A 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()

A 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

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.

Code:

#Group(), Groups() &amp; Groupdict() hackerrank Solution

import re
m=input()

al=r'([a-z0-9A-Z])\1+'
a=re.search(al,m)
if a:
    print(a.group(1))
else:
    print(-1)

#Group(), Groups() &amp; Groupdict() hackerrank Solution

Disclaimer: This problem is originally created and published by HackerRank, we only provide solutions to this problem. Hence, doesn’t guarantee the truthfulness of the problem. This is only for information purposes.