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Python | Get the starting index for all occurrences of given substring

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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))


Output:

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))


Output:

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))


Output:

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))


Output

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]


Output

[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)



Last Updated : 24 Mar, 2023
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