This particular article is indeed a very useful one for Machine Learning enthusiast as it solves a good problem for them. In Machine Learning we generally encounter this issue of getting a particular string in huge amount of data and handling that sometimes becomes a tedious task. Lets discuss certain way outs to solve this problem. 

Method #1: Using remove() This particular method is quite naive and not recommended to use, but is indeed a method to perform this task. remove() generally removes the first occurrence of K string and we keep iterating this process until no K string is found in list. 

# Python 3 code to demonstrate
# Remove K String from String List
# using remove()

# initializing list
test_list = ["bad", "GeeksforGeeks", "bad", "is", "best", "bad"]

# Printing original list
print("Original list is : " + str(test_list))

# initializing K
K = "bad"

# using remove() to
# Remove K String from String List
while(K in test_list):
    test_list.remove(K)

# Printing modified list
print("Modified list is : " + str(test_list))

Original list is : ['bad', 'GeeksforGeeks', 'bad', 'is', 'best', 'bad']
Modified list is : ['GeeksforGeeks', 'is', 'best']

Time Complexity: O(n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(1), constant extra space is required

Method #2: Using List Comprehension More concise and better approach to remove all the K strings, it just checks if the string is not K and re-makes the list with all strings that are not K. 

# Python 3 code to demonstrate
# Remove K String from String List
# using list comprehension

# initializing list
test_list = ["bad", "GeeksforGeeks", "bad", "is", "best", "bad"]

# Printing original list
print("Original list is : " + str(test_list))

# initializing K
K = "bad"

# using list comprehension to
# Remove K String from String List
test_list = [i for i in test_list if i != K]

# Printing modified list
print("Modified list is : " + str(test_list))

Original list is : ['bad', 'GeeksforGeeks', 'bad', 'is', 'best', 'bad']
Modified list is : ['GeeksforGeeks', 'is', 'best']

Time Complexity: O(n)
Auxiliary Space: O(n), where n is length of list.

Method #3 : Using join(),replace(),split() and remove() methods

# Python 3 code to demonstrate
# Remove K String from String List

# initializing list
test_list = ["bad", "GeeksforGeeks", "bad", "is", "best", "bad"]

# Printing original list
print ("Original list is : " + str(test_list))

# initializing K
K = "bad"

x="-".join(test_list)
x=x.replace(K,"")
a=x.split("-")
while("" in a ):
    a.remove("")
# Printing modified list
print ("Modified list is : " + str(a))

Original list is : ['bad', 'GeeksforGeeks', 'bad', 'is', 'best', 'bad']
Modified list is : ['GeeksforGeeks', 'is', 'best']

Here's another approach to removing a string 'K' from a string list, using a filter:

# Python 3 code to demonstrate
# Remove K String from String List
# using filter()
 
# initializing list
test_list = ["bad", "GeeksforGeeks", "bad", "is", "best", "bad"]
 
# Printing original list
print("Original list is : " + str(test_list))
 
# initializing K
K = "bad"
 
# using filter to Remove K String from String List
test_list = list(filter(lambda x: x!=K, test_list))
 
# Printing modified list
print("Modified list is : " + str(test_list))

Original list is : ['bad', 'GeeksforGeeks', 'bad', 'is', 'best', 'bad']
Modified list is : ['GeeksforGeeks', 'is', 'best']

This approach uses the filter() function to create a filtered list of elements from test_list that are not equal to K. The list() function is then used to convert the filtered list into a list.

Time complexity: O(n), where n is the number of elements in the list test_list.

Space complexity: O(n), since a new list is created.