How to remove text inside brackets in Python?
In this article, we will learn how to remove content inside brackets without removing brackets in python.
Input: (hai)geeks Output: ()geeks Input: (geeks)for(geeks) Output: ()for()
We can remove content inside brackets without removing brackets in 2 methods, one of them is to use the inbuilt methods from the re library and the second method is to implement this functionality by iterating the string using a for loop
Method 1: We will use sub() method of re library (regular expressions).
sub(): The functionality of sub() method is that it will find the specific pattern and replace it with some string.
This method will find the substring which is present in the brackets or parenthesis and replace it with empty brackets.
- Import the re library
- Now find the sub-string present in the brackets and replace with () using sub() method.
- We need to pass the sub() method with 2 arguments those are pattern and string to be replaced with.
- Print the string.
In the below code \(.*?\) represents the regular expression for finding the brackets containing some content. The brackets () have some special meaning in regular expression in python so Backlash \ is used to escape that meaning.
Time complexity: O(2^m + n).
Where m is the size of the regex, and n is the size of the string. Here the sub() method will take 2^m time to find the pattern using the regex and O(n) to iterate through the string and replace with the “()”.
Method 2: In this method, we will iterate through the string and if the character that is being iterated is not present in between the parenthesis then the character will be added to the resultant string.
If there is an open or closed parenthesis present in the string then the parenthesis will be added to the resultant string but the string in between them is not added to the resultant string.
- Iterate the loop for each character in the string.
- If a ‘(‘ or ‘)’ appears in the string we will add it to the result string.
- If the parenthesis number in the string is zero then add the character to the result string.
- Here if the parenthesis number is greater than zero it indicates that the current character which is being iterated is present in between two parentheses
- Print the result string.
Time complexity : O(n).
Here n is the length of the string. In the code, we are iterating through the string and appending the content outside of the parenthesis so it would only take the time O(n).
Method 3 : Using replace(),split() and join() methods
Time Complexity : O(N)
Auxiliary Space : O(N)