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

Validating UID

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554 Discussions

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  • rohitambasta_ii1
     + 0 comments
    T= int(input())

for _ in range(T):
    criteria = []
    user_id_str= input()
    user_id = [i for i in user_id_str]
    
    #1. At least 2 uppercase English alphabet 
    criteria.append((len([element.isupper() for element in user_id if element.isupper() is True])) >= 2)
    
    #2. At least 3 digits
    criteria.append((len([element.isdigit() for element in user_id if element.isdigit() is True])) >= 3)   
       
    #3. only contain alphanumeric
    criteria.append(user_id_str.isalnum())
    
    #4. No character should repeat
    criteria.append((len(user_id)) == (len(set(user_id))))
    
    #5. Must be exactly 10 characters 
    criteria.append(len(user_id) == 10)
    
    print("Valid" if all(criteria) else "Invalid")

  • satyamroy173
     + 0 comments

    import re

    uid = list T = int(input()) pattern = r'^(?=(?:.[A-Z]){2,})(?=(?:.\d){3,})[A-Za-z\d]+$'

    for _ in range(T): # print(T) id = input() if len(set(id))==10: if re.match(pattern, id): print("Valid") else: print("Invalid") else: print("Invalid")

  • maluisamrok
     + 0 comments
    import re
n= int(input())
valido =True
     
for _ in range(n):
    UID =input();
    if len(UID)==10:
        valido = True
    else:
        valido =False
    
    if valido  and UID.isalnum():
        valido =True
    else:
        valido = False
    if valido and len(re.findall(r'[A-Z]',UID))>=2:
        valido = True
    else:
        valido =False
    if valido and len(re.findall(r'[0-9]',UID))>=3:
        valido = True
    else:
        valido =False
    
    for c in UID:
        if len(re.findall(c,UID))!=1:
            valido=False
            break
        
        
    print("Valid" if valido else "Invalid")

  • 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)