HackerRank Validating UID solution in python

In this Validating UID problem, ABCXYZ company has up to 100 employees. The company decides to create a unique identification number (UID) for each of its employees. The company has assigned you the task of validating all the randomly generated UIDs.

Problem solution in Python 2 programming.

import re
for i in range(int(raw_input())):
    N = raw_input().strip()
    if N.isalnum() and len(N) == 10:
        if bool(re.search(r'(.*[A-Z]){2,}',N)) and bool(re.search(r'(.*[0-9]){3,}',N)):
            if re.search(r'.*(.).*\1+.*',N):
                print 'Invalid'
            else:
                print 'Valid'    
        else:
            print 'Invalid'
    else:
        print 'Invalid'

Problem solution in Python 3 programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT
import re

for _ in range(int(input())):
    u = ''.join(sorted(input()))
    try:
        assert re.search(r'[A-Z]{2}', u)
        assert re.search(r'\d\d\d', u)
        assert not re.search(r'[^a-zA-Z0-9]', u)
        assert not re.search(r'(.)\1', u)
        assert len(u) == 10
    except:
        print('Invalid')
    else:
        print('Valid')

Problem solution in pypy programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT

ids = []
line_num = int(raw_input())
while line_num:
  ids.append(raw_input());
  line_num -= 1;

import re

for id in ids:
  chars_count = {char:id.count(char) for char in id}
  upper = 0
  not_alnum = 0;
  dig = 0;
  for ch in id:
    if ch.isupper(): upper += 1
    if not ch.isalnum(): not_alnum = 1;
    if ch.isdigit(): dig += 1;

  if len(id) != 10: print 'Invalid'
  elif len(set(chars_count.values())) != 1: print 'Invalid'
  elif upper < 2: print 'Invalid'
  elif dig < 3: print 'Invalid'
  elif not_alnum: print 'Invalid'
  else: print 'Valid'

Problem solution in pypy3 programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT
import re

def valid_uid(uid):
    if len(re.findall(r"[A-Z]", uid)) < 2 or \
            len(re.findall(r"[0-9]", uid)) < 3 or \
            not re.match(r"[A-Za-z0-9]{10}$", uid) or \
            len(set(uid)) != len(uid):
        return False
    
    return True
    
n = int(input())
for _ in range(n):
    uid = input()
    if valid_uid(uid):
        print("Valid")
    else:
        print("Invalid")