Python | Replace punctuations with K

Sometimes, while working with Python Strings, we have a problem in which we need to perform the replacement of punctuations in String with specific characters. This can have applications in many domains such as day-day programming. Let’s discuss certain ways in which this task can be performed.

Method #1: Using string.punctuation + replace() The combination of the above functions can be used to solve this problem. In this, we extract all punctuations using punctuation and perform a replacement of the character desired using replace(). 

The original string is : geeksforgeeks, is : best for ; geeks!!
The strings after replacement : geeksforgeeks* is * best for * geeks**

Method #2: Using regex This problem can be solved using regex. In this, we employ a regex to substitute for all punctuations. 

The original string is : geeksforgeeks, is : best for ; geeks!!
The strings after replacement : geeksforgeeks* is * best for * geeks**

Time Complexity: O(N)
Space Complexity: O(N)

Method #3: Using translate() method replace punctuations with K:

1. Initialize the input string “est_str.
2. Initialize the replace character repl_char.
3. Create a translator using str.maketrans() method with two arguments the first argument is string.punctuation(), which is the list of all punctuations in the string module and The second argument is the “*” character multiplied by the length of the first argument, which will be the replacement character.
4. Use the translate() method to replace all the punctuation with the replacement character.
5. Store the translated string in “test_str” and print the string after replacement.

Below is the implementation of the above approach:

The original string is : geeksforgeeks, is : best for ; geeks !!
The strings after replacement : geeksforgeeks* is * best for * geeks **

Time Complexity: O(N)
Space Complexity: O(N)

Method #4: using the join() method:

• The string test_str is initialized
• The replace character repl_char is initialized 
• The string of all possible punctuation marks is initialized using the punctuation module of the string library.
• The join() method is used to replace each punctuation mark in the string test_str with the replacement character. 
• The join() method is used in combination with a ternary operator which checks if each character in test_str is a punctuation mark or not. 
• If it is a punctuation mark, then the replacement character is used, otherwise the original character is used. The result of this operation is a string with all the punctuation marks replaced with the replacement character.
• The resulting string is printed using print() function.

The original string is : geeksforgeeks, is : best for ; geeks !!
The string after replacement : geeksforgeeks* is * best for * geeks **

The time complexity of this approach is O(n), where n is the length of the input string.
The space complexity of this approach is O(n), where n is the length of the input string

Method 5: Using a loop to iterate through each character in the string and checking if it is a punctuation character.

Step-by-step approach

• Initialize the replace character repl_char to ‘*’.
• Initialize the string of punctuation characters punctuations to ”’!()-[]{};:'”,<>./?@#$%^&*_~”’.
• Initialize an empty string res to store the modified string.
• Use a loop to iterate through each character in the input string test_str.
• Check if the current character is a punctuation character by using the in operator to check if it exists in the punctuations string.
• If the current character is a punctuation character, then append the repl_char to the res string.
• If the current character is not a punctuation character, then append the current character to the res string.
• After iterating through all the characters in the input string, print the modified string using the print() function.

The original string is : geeksforgeeks, is : best for ; geeks !!
The string after replacement : geeksforgeeks* is * best for * geeks **

Time complexity: O(n) where n is the length of the input string since we iterate through each character in the string once.
Auxiliary space: O(n) since we use a new string to store the modified string which can be of the same length as the input string if all characters are punctuation characters.

Method 6: Using  recursive implementation:

1. Define a string test_str.
2. Initialize the replace character repl_char to *.
3. Define a string of punctuations punctuations.
4. Initialize an empty string res.
5. Iterate over each character in the test_str.
6. If the character is a punctuation mark, append repl_char to res, otherwise append the character to res.
7. Print the original string and the modified string.

The original string is : geeksforgeeks, is : best for ; geeks !!
The string after replacement : geeksforgeeks* is * best for * geeks **

The time complexity: O(n), where n is the length of the input string test_str. This is because we iterate over each character in the string exactly once, and the time taken to perform the other operations is constant.
The space complexity: O(n), because we are creating a new string res of the same length as the input string to store the modified string.

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• Python
• Python Programs
• Python string-programs

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