- Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
- You must implement a solution with a linear runtime complexity and use only constant extra space.
Input: nums = [2,2,1]
Output: 1
Input: nums = [2,2,1]
Output: 1
Input: n = 3
Output: ["1","2","Fizz"]
>>> 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')
>>> import re
>>> m = re.search(r'\d+','1234')
>>> m.end()
4
>>> m.start()
0
>>> import re
>>> re.findall(r'\w','http://www.hackerrank.com/')
['h', 't', 't', 'p', 'w', 'w', 'w', 'h', 'a', 'c', 'k', 'e', 'r', 'r', 'a', 'n', 'k', 'c', 'o', 'm']
There are a total of [occupation_count] [occupation]s.
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