Given a string and ranges list, remove all the characters that occur in ranges.
Input : test_str = ‘neveropen is best for Lazyroar’, range_list = [(3, 6), (7, 10)]
Output : Lazyroar is best for Lazyroar
Explanation: The required ranges removed.Input : test_str = ‘neveropen is best for Lazyroar’, range_list = [(3, 6)]
Output : georLazyroar is best for Lazyroar
Explanation : The required ranges removed.
Method #1: Using loop
In this, we check for each range, and remake string, considering the index doesn’t lie in range checking using conditional statements.
Python3
# Python3 code to demonstrate working of # Remove index ranges from String # Using loop# initializing stringstest_str1 = 'neveropen is best for Lazyroar'# printing original stringprint("The original string 1 is : " + str(test_str1))# initializing ranges list range_list = [(3, 6), (7, 10), (14, 17)]res = ""for idx, chr in enumerate(test_str1): for strt_idx, end_idx in range_list: # checking for ranges and appending if strt_idx <= idx + 1 <= end_idx: break else: res += chr# printing result print("The reconstructed string : " + str(res)) |
The original string 1 is : neveropen is best for Lazyroar The reconstructed string : Lazyroarbest for Lazyroar
Method #2 : Using any() + list comprehension + join()
In this, we perform the task of checking for indices for strings using any() and list comprehension is used to reconstruct string accordingly.
Python3
# Python3 code to demonstrate working of # Remove index ranges from String # Using any() + list comprehension + join()# initializing stringstest_str1 = 'neveropen is best for Lazyroar'# printing original stringprint("The original string 1 is : " + str(test_str1))# initializing ranges list range_list = [(3, 6), (7, 10), (14, 17)]# using any() to check for strings in index rangesres = ''.join(chr for idx, chr in enumerate(test_str1, 1) if not any(strt_idx <= idx <= end_idx for strt_idx, end_idx in range_list))# printing result print("The reconstructed string : " + str(res)) |
The original string 1 is : neveropen is best for Lazyroar The reconstructed string : Lazyroarbest for Lazyroar
Time Complexity: O(n2 )
Auxiliary Space: O(n)
Method #3: Using itertools.compress()
- Import the itertools module for compress operations.
- Initialize an empty string res.
- Initialize a list indices of the same length as test_str1, with all values set to True.
Iterate over the range_list, and for each (start_idx, end_idx) in the list:- Update the corresponding indices in indices to False for the given range.
- Use itertools.compress() function to filter out the characters from test_str1 using the indices list and join them to form the reconstructed string.
- Print the reconstructed string res.
Python3
# Python3 code to demonstrate working of # Remove index ranges from String # Using itertools.compress()import itertools# initializing stringstest_str1 = 'neveropen is best for Lazyroar'# printing original stringprint("The original string 1 is : " + str(test_str1))# initializing ranges list range_list = [(3, 6), (7, 10), (14, 17)]res = ""indices = [True] * len(test_str1)for start_idx, end_idx in range_list: for i in range(start_idx-1, end_idx): indices[i] = Falseres = ''.join(itertools.compress(test_str1, indices))# printing result print("The reconstructed string : " + str(res)) |
The original string 1 is : neveropen is best for Lazyroar The reconstructed string : Lazyroarbest for Lazyroar
Time complexity: O(n), where n is the length of the input string test_str1.
Auxiliary space: O(n), where n is the length of the input string test_str1.
