Python | Remove given character from Strings list

Sometimes, while working with Python list, we can have a problem in which we need to remove a particular character from each string from list. This kind of application can come in many domains. Let’s discuss certain ways to solve this problem. 

Method #1 : Using replace() + enumerate() + loop 

This is brute force way in which this problem can be solved. In this, we iterate through each string and replace specified character with empty string to perform removal. 

Step-by-step approach:

1. Initialize a character named char which is to be removed from the strings in the list.
2. Iterate over the list using a for loop and enumerate() function to get the index and corresponding element in each iteration.
3. For each iteration, replace the character ‘s’ with an empty string ” in the corresponding element using the replace() method and update the list with the modified element at the same index using the index obtained from enumerate().
4. Print the modified list of strings.

Below is the implementation of the above approach:

The original list : ['gfg', 'is', 'best', 'for', 'Lazyroar']
The list after removal of character : ['gfg', 'i', 'bet', 'for', 'geek']

Time complexity: O(n*m), where n is the length of the input list and m is the average length of the strings in the list. 
Auxiliary space: O(1). The algorithm modifies the original list in place and does not create any new data structures. 

Method #2: Using list comprehension + replace() 

This task can also be performed using the above functionalities. In this, we offer a one-liner solution to this problem compacting the code using similar method as above. 

The original list : ['gfg', 'is', 'best', 'for', 'Lazyroar']
The list after removal of character : ['gfg', 'i', 'bet', 'for', 'geek']

Time complexity: O(n*m), where n is the length of the original list and m is the length of the longest string in the list

Auxiliary space: O(n*m).

Method #3: Using map(),lambda functions.

The original list : ['gfg', 'is', 'best', 'for', 'Lazyroar']
The list after removal of character : ['gfg', 'i', 'bet', 'for', 'geek']

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

Method#4: Using Recursive method.

Algorithm:

1. Define a function to remove the character from the given list of strings.
2. The function takes two inputs, the list of strings lst and the character to be removed char.
3. Check if the list is empty. If yes, return an empty list.
4. If the list is not empty, remove the character from the first string in the list using the replace method, and add it to a list.
5. Recursively call the function with the rest of the list, and add the result to the list created in the previous step.
6. Return the modified list of strings.

The original list : ['gfg', 'is', 'best', 'for', 'Lazyroar']
The list after removal of character : ['gfg', 'i', 'bet', 'for', 'geek']

Time Complexity:
The time complexity of this function is O(n * m), where n is the number of strings in the list and m is the maximum length of any string in the list. This is because for each string, the replace method scans the entire string character by character, which takes up to m time. We need to perform this operation for each of the n strings in the list, resulting in a time complexity of O(n * m).

Auxiliary Space:
The space complexity of this function is O(n * m), where n is the number of strings in the list and m is the maximum length of any string in the list. This is because we create a new list to store the modified strings, and the size of each modified string could be up to m characters. We need to create this list for each of the n strings in the list, resulting in a space complexity of O(n * m).

Method #6: Using regex

1. Import the re module.
2. Define the function remove_char(lst, char) that takes a list of strings and a character as arguments.
3. Use the re.sub() function to replace the specified character in each string in the list with an empty string.
4. Return the modified list.

The original list : ['gfg', 'is', 'best', 'for', 'Lazyroar']
The list after removal of character : ['gfg', 'i', 'bet', 'for', 'geek']

Time complexity: O(nm), where n is the number of strings in the list and m is the average length of the strings. The re.sub() function has a time complexity of O(m) for each string in the list.

Auxiliary space: O(nm), where n is the number of strings in the list and m is the average length of the strings. This is the space required to store the modified strings in the new list.

Method #7: Using reduce():

Algorithm:

1. Import the “reduce” function from the “functools” module and the “re” module for regular expressions.
2. Define the “remove_char” function that takes a list of strings and a character as input.
3. Use the “reduce” function to iterate over each string in the list and apply the “re.sub()” function to remove the specified character.
4. Append the modified string to a new list.
5. Return the new list with the modified strings.
6. Initialize a list of strings “test_list”.
7. Print the original list.
8. Initialize a character “char”.
9. Call the “remove_char” function with the “test_list” and “char” as arguments.
10. Print the modified list.

The original list : ['gfg', 'is', 'best', 'for', 'Lazyroar']
The list after removal of character : ['gfg', 'i', 'bet', 'for', 'geek']

Time complexity:
The time complexity of the “remove_char” function is O(nm), where “n” is the length of the input list and “m” is the length of the longest string in the list. The “reduce” function takes O(n) time to iterate over each string in the list, and the “re.sub()” function takes O(m) time to remove the specified character from each string.

Space complexity:
The space complexity of the “remove_char” function is O(nm), where “n” is the length of the input list and “m” is the length of the longest string in the list. The function creates a new list to store the modified strings, which takes up O(nm) space.