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

Validating UID

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556 Discussions

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  • maxence_qnt
     + 0 comments

    I have used regex to solve this problem, and some python logics.

    I have created 3 differents patterns (1 for each condition, except the last one which check at the same time if the UID is exactly 10 characters).

    first condition pattern check if 2 uppercase (at least) are present or not.

    Note : ?. is a lookaround operators and it means anything before an uppercase. So "....A....". This format must be repeated at least twice : {2,}

    Exact same logics with digits 0-9

    And the last pattern, I think, is logic to understand, we only want a-zA-Z0-9 characters exactly 10 times.

    Then, we just have to check condition by condition.

    Here my code, I am pretty sure there is a simple way or more concise way to solve this problem, but, it works and the logics sounds good :

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

T = int(input())

UID_list = list()

for _ in range(T):
    current_UID = input()
    UID_list.append(current_UID)
    
# Patterns
first_condition_pattern = r'^(?=(?:.*[A-Z]){2,}).*$'

second_condition_pattern = r"^(?=(?:.*[0-9]){3,}).*$"

third_condition_pattern = r"[a-zA-Z0-9]{10}"

for element in UID_list:
    
    if len(set(element)) != len(list(element)):
        #print("Repeat character")
        print("Invalid")

    elif bool(re.match(first_condition_pattern, element)) == False:
        #print("Not 2 upper case minimum")
        print("Invalid")
    
    elif bool(re.match(second_condition_pattern, element)) == False:
        #print("Not 3 digits minimum")
        print("Invalid")
    
    elif bool(re.match(third_condition_pattern, element)) == False:
        #print("Not only alphanumeric character or not exactly 10 characters")
        print("Invalid")
    
    else:
        print("Valid")

  • pithambar20
     + 0 comments
    N = int(input())


l1 = []

for i in range(N):
    inp = input()
    l1.append([ch for ch in inp])


l2 = [set(item) for item in l1]


upperCase = 0
digits = 0

for i in range(N):
    
    for ch in l1[i]:    
        if ch.isupper():
            upperCase+=1
        if ch.isdigit():
            digits += 1
            
    if all([len(l1[i]) == len(l2[i]) and len(l1[i])==10,upperCase>=2,digits>=3]):
        print("Valid")
    else:
        print("Invalid")
        
    upperCase = 0
    digits = 0

  • 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")