Python – Check if Elements delimited by K

Given a String, check if each segment is delimited by  K.

Input : test_str = ’72!45!geeks!best’, K = ‘!’ 
Output : True 
Explanation : All numerics and alphabets separated by delim border. 

Input : test_str = ’72!45geeks!best’, K = ‘!’ 
Output : False 
Explanation : No separation between 45 and geeks.

Method 1: Using isdigit() + isalpha() + loop

This is way in which this task can be performed. In this, we perform task of checking for alphabets and digits segments using isalpha() and isdigit(). Presence of any element, not entirely number or alphabet, is termed to non-delimited by K, and it remains unresolved during split().

Python3

# Python3 code to demonstrate working of
# Check if Elements delimited by K
# Using isdigit() + isalpha() + loop

# initializing string
test_str = '72@45@geeks@best'

# printing original string
print("The original string is : " + str(test_str))

# initializing splt_chr
K = "@"

res = True

# splitting elements
temp = test_str.split(K)

for ele in temp:

    # checking for non-alpha or non-digit
    if len(ele) > 1 and not ele.isdigit() and not ele.isalpha():
        res = False
        break

# printing result
print("Are all delimited by K : " + str(res))

Output

The original string is : 72@45@geeks@best
Are all delimited by K : True

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #2: Using the re module to split the string and check for elements

Step-by-step algorithm:

• Split the given string using the delimiter K using re.split() method and store in a temporary variable ‘temp’.
• Check if each element in ‘temp’ is either a single character or consists of only alphabets or only digits using all() method.
• Print the final result.

Python3

import re
# initializing string
test_str = '72@45@geeks@best'
# initializing splt_chr
K = "@"
# printing original string
print("The original string is : " + str(test_str))

temp = re.split(re.escape(K), test_str)
res = all(len(ele) == 1 or ele.isdigit() or ele.isalpha() for ele in temp)

# printing result
print("Are all delimited by K : " + str(res))

Output

The original string is : 72@45@geeks@best
Are all delimited by K : True

Time Complexity: O(n), where n is the length of the given string test_str since we need to iterate over each element in ‘temp’.
Auxiliary Space: O(n), since we are storing the split string in a list ‘temp’ of size n.

Method #3: Using all() and generator expression

• Split the string test_str based on the delimiter K using the split() method and generate a list of substrings.
• Check if each element of the list generated in the previous step is valid by iterating over it and applying the following conditions:
  • len(ele) == 1 checks if the element has a length of 1.
  • ele.isalpha() checks if the element is an alphabet.
  • ele.isdigit() checks if the element is a digit.
• Use the all() method to check if all the elements of the list satisfy the above conditions and assign the result to a variable res.
• Print the result of whether all elements are delimited by the specified delimiter or not using the print() statement.
  

Python3

# initializing string
test_str = '72@45@geeks@best'

# initializing splt_chr
K = "@"

# printing original string
print("The original string is : " + str(test_str))

# split the string based on delimiter
# and check if each element is valid
res = all(len(ele) == 1 or ele.isalpha() or ele.isdigit()
          for ele in test_str.split(K))

# print the result
print("Are all delimited by K: " + str(res))

Output

The original string is : 72@45@geeks@best
Are all delimited by K: True

Time Complexity: O(n), where n is the length of the given string test_str since we need to iterate over each element in ‘temp’.
Auxiliary Space: O(n), since we are storing the split string in a list ‘

Method #4:  Manual Splitting and Validation Method

1. Initialize a boolean variable valid to True.
2. Initialize an empty string current to store the current sub-string.
3. Loop through each character in the input string.
4. If the current character is the delimiter character K, check if the current sub-string is valid by either checking its length or using the isalpha() and isdigit() methods. If it’s not valid, set the valid variable to False.
5. If the current character is not the delimiter character K, append it to the current sub-string.
6. After the loop is done, check the validity of the last sub-string by checking its length or using the isalpha() and isdigit() methods. If it’s not valid, set the valid variable to False.
7. Print the value of the valid variable

Python3

# initializing string
test_str = '72@45@geeks@best'

# initializing splt_chr
K = "@"

# printing original string
print("The original string is : " + str(test_str))

# initialize variables
valid = True
current = ""

# loop through each character in the input string
for c in test_str:
    if c == K: # if delimiter
        if len(current) == 0 or not (current.isalpha() or current.isdigit()):
            valid = False
            break
        current = ""
    else: # if not delimiter
        current += c

# check last sub-string
if len(current) == 0 or not (current.isalpha() or current.isdigit()):
    valid = False

# print the result
print("Are all delimited by K: " + str(valid))

Output

The original string is : 72@45@geeks@best
Are all delimited by K: True

Time complexity: O(n) where n is the length of the input string.
Auxiliary space: O(1) as we are using constant amount of space for storing variables.

Method #5: Using the split() method and the isalnum() method:

Python3

# initializing string
test_str = '72@45@geeks@best'

# initializing delimiter
delimiter = '@'

# printing original string
print("The original string is : " + str(test_str))

# split the string using the delimiter
split_str = test_str.split(delimiter)

# initialize variables
valid = True

# loop through each substring and check if it contains only alphanumeric characters
for s in split_str:
    if not s.isalnum():
        valid = False
        break

# print the result
print("Are all delimited by K: " + str(valid))

Output

The original string is : 72@45@geeks@best
Are all delimited by K: True

Time complexity: O(n), where n is the length of the input string.
Auxiliary space: O(1) as we are using constant amount of space for storing variables.