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  1. Prepare
  2. Python
  3. Regex and Parsing
  4. Validating UID
  5. Discussions

Validating UID

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551 Discussions

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  • jimhark
     + 0 comments

    After reviewing the disdcussions here, I realized the code I just posted should have the following improvements:

    • Don't use a counter to detect character repeats, set is much better
    • isalnum() works on string, don't go character by character
    • In this case, generater expressions are better than list comprehensions

    Here is the updated code:

    # A valid UID must:
#
#   * Be exactly 10 characters in length
#   * Only contain alphanumeric characters (a-z, A-Z, 0-9)
#   * Contain at least 2 uppercase English alphabet characters
#   * Contain at least 3 digits (0-9)
#   * Not repeat any character

def validate_uid(uid):
    if len(uid) != 10:
        return False

    if not uid.isalnum():
        return False

    if sum(c.isupper() for c in uid) < 2:
        return False

    if sum(c.isdigit() for c in uid) < 3:
        return False

    if len(set(uid)) != len(uid):
        return False

    return True

t = int(input())

for _ in range(t):
    uid = input()
    is_valid = validate_uid(uid)

    message = 'Valid' if is_valid else 'Invalid'
    print(message)

  • jimhark
     + 0 comments

    I've had too many regex solutions in a row, so I decided to go a different way for this. I tried to focus on readability (and debugability).

    import collections

# A valid UID must:
#
#   * Be exactly 10 characters in length
#   * Only contain alphanumeric characters (a-z, A-Z, 0-9)
#   * Contain at least 2 uppercase English alphabet characters
#   * Contain at least 3 digits (0-9)
#   * Not repeat any character

def validate_uid(uid):
    if len(uid) != 10:
        return False

    if not all( [c.isalnum() for c in uid] ):
        return False

    if sum([c.isupper() for c in uid]) < 2:
        return False

    if sum([c.isdigit() for c in uid]) < 3:
        return False

    if max(collections.Counter(uid).values()) > 1:
        return False

    return True

t = int(input())

for _ in range(t):
    uid = input()
    is_valid = validate_uid(uid)

    message = 'Valid' if is_valid else 'Invalid'
    print(message)

  • yashdeora98294
     + 0 comments

    Here is HackerRank Validating UID in Python solution - https://programmingoneonone.com/hackerrank-validating-uid-solution-in-python.html

  • singhshrajan
     + 0 comments
    1. import re
      1. def check(uid):
    3. if len(uid)!=len(set(uid)):
    4. return "Invalid"
    5. else:
    6. match = re.match(r'^(?=(?:.[A-Z]){2,})(?=(?:.\d){3,})[A-Za-z0-9]{10}$',uid)
    7. if match:
    8. return "Valid"
    9. else:
    10. return "Invalid"
      1. for i in range(int(input())):
    12. print(check(input()))

  • illya_muravsky
     + 1 comment
    def uid_validator(uid):
	uid = uid.strip()
	return (
		'Valid' if (
			re.fullmatch(r'^[0-9a-zA-Z]¨{10}', uid)
			and len(re.findall(r'[A-Z], uid) >=2)
			and len(re.findall(r'[0-9], uid) >= 3)
			and len(set(uid) == 10
		) else 'Invalid
	)