Python | Get the starting index for all occurrences of given substring
Given a string and a substring, the task is to find out the starting index for all the occurrences of a given substring in a string. Let’s discuss a few methods to solve the given task.
Method #1: Using Naive Method
Python3
# Python3 code to demonstrate # to find all occurrences of substring in # a string # Initialising string ini_string = 'xbzefdgstbzefzexezef' # Initialising sub-string sub_string = 'zef' # Printing initial string and sub-string print ( "initial_strings : " , ini_string, "\nsubstring : " , sub_string) res = [] flag = 0 k = 0 # Finding all occurrences of substring # in a string using Naive method for i in range ( 0 , len (ini_string)): k = i flag = 0 for j in range ( 0 , len (sub_string)): if ini_string[k] ! = sub_string[j]: flag = 1 if flag: break k = k + 1 if flag = = 0 : res.append(i) # printing result( print ( "resultant positions" , str (res)) |
initial_strings : xbzefdgstbzefzexezef substring : zef resultant positions [2, 10, 17]
Time Complexity: O(n2)
Auxiliary Space O(n)
Method #2: Using list comprehension
Python3
# Python3 code to demonstrate # to find all occurrences of substring in # a string # Initialising string ini_string = 'xbzefdgstbzefzexezef' # Initialising sub-string sub_string = 'zef' # Printing initial string and sub-string print ( "initial_strings : " , ini_string, "\nsubstring : " , sub_string) res = [] # Finding all occurrences of substring # in a string using list comprehension res = [i for i in range ( len (ini_string)) if ini_string.startswith(sub_string, i)] # printing result( print ( "resultant positions" , str (res)) |
initial_strings : xbzefdgstbzefzexezef substring : zef resultant positions [2, 10, 17]
Time Complexity: O(n2)
Auxiliary Space O(n)
Method #3: Using regex
Python3
# Python3 code to demonstrate # to find all occurrences of substring in # a string import re # Initialising string ini_string = 'xbzefdgstbzefzexezef' # Initialising sub-string sub_string = 'zef' # Printing initial string and sub-string print ( "initial_strings : " , ini_string, "\nsubstring : " , sub_string) res = [] # Finding all occurrences of substring # in a string using re.finditer res = [m.start() for m in re.finditer(sub_string, ini_string)] # printing result( print ( "resultant positions" , str (res)) |
initial_strings : xbzefdgstbzefzexezef substring : zef resultant positions [2, 10, 17]
Time Complexity: O(n2)
Auxiliary Space: O(n)
Method #4 : Using find() and replace() methods
Python3
# Python3 code to demonstrate # to find all occurrences of substring in # a string # Initialising string ini_string = 'xbzefdgstbzefzexezef' # Initialising sub-string sub_string = 'zef' # Printing initial string and sub-string print ( "initial_strings : " , ini_string, "\nsubstring : " , sub_string) res = [] while (ini_string.find(sub_string) ! = - 1 ): res.append(ini_string.find(sub_string)) ini_string = ini_string.replace(sub_string, "*" * len (sub_string), 1 ) # printing result( print ( "resultant positions" , str (res)) |
initial_strings : xbzefdgstbzefzexezef substring : zef resultant positions [2, 10, 17]
Time Complexity: O(n2)
Auxiliary Space: O(n)
Using str.index() in a loop:
Approach:
In this example, the string variable contains the string we want to search in, and the substring variable contains the substring we want to find. We pass these variables as arguments to the get_substring_indices() function and store the result in the indices variable. Finally, we print the indices variable to see the starting index for all occurrences of the given substring.
Python3
def get_substring_indices(string, substring): indices = [] try : index = string.index(substring) while index ! = - 1 : indices.append(index) index = string.index(substring, index + 1 ) except ValueError: pass return indices string = "hello world, world is beautiful" substring = "world" indices = get_substring_indices(string, substring) print (indices) # Output: [6, 18] |
[6, 13]
Time Complexity: O(n*m), where n is the length of the string and m is the length of the substring
Auxiliary Space: O(1)