Python | Ways to find nth occurrence of substring in a string

Given a string and a substring, write a Python program to find the n^th occurrence of the string. Let’s discuss a few methods to solve the given task. 

Get Nth occurrence of a substring in a String using regex

Here, we find the index of the ‘ab’ character in the 4th position using the regex re.finditer()

Python3

import re

# Initialising values
ini_str = "abababababab"
substr = "ab"
occurrence = 4

# Finding nth occurrence of substring
inilist = [m.start() for m in re.finditer(r"ab", ini_str)]
if len(inilist)>= 4:

  # Printing result
  print ("Nth occurrence of substring at", inilist[occurrence-1])
else:
  print ("No {} occurrence of substring lies in given string".format(occurrence))

Output:

Nth occurrence of substring at 6

Get Nth occurrence of a substring in a String using find() method

Here, we find the index of the ‘ab’ character in the 4th position using the str.find().

Python3

# Initialising values
ini_str = "abababababab"
sub_str = "ab"
occurrence = 4

# Finding nth occurrence of substring
val = -1
for i in range(0, occurrence):
  val = ini_str.find(sub_str, val + 1)

# Printing nth occurrence
print ("Nth occurrence is at", val)

Output:

Nth occurrence is at 6

Get Nth occurrence of a substring in a String using startswith() 

Here, we find the index of the ‘ab’ character in the 4th position using the str.startwith().

Python3

# Initialising values
ini_str = "abababababab"
substr = "ab"
occurrence = 4


# Finding nth occurrence of substring
inilist = [i for i in range(0, len(ini_str))
      if ini_str[i:].startswith(substr)]

if len(inilist)>= 4:

  # Printing result
  print ("Nth occurrence of substring at", inilist[occurrence-1])
else:
  print ("No {} occurrence of substring lies in given string".format(occurrence))

Output:

Nth occurrence of substring at 6

Get Nth occurrence of a substring in a String using split()

Here, we find the index of the ‘ab’ character in the 4th position using the split().

Python3

def solve(s, str, n):
    sep = s.split(str, n)
    if len(sep) <= n:
        return -1
    return len(s) - len(sep[-1]) - len(str)

print('length: ', len('dellGeeks asusfor geeksforgeeksdell'))
print("position", solve('dellGeeks asusfor geeksforgeeksdell', 'dell', 2))

Output:

length:  35
position 31

The time complexity of the given code is O(n), where n is the length of the input string s.

The space complexity of the code is O(n), where n is the length of the input string s. 

Using the count() method:

Approach:

Use the count() method to find the total number of occurrences of the substring in the given string.
If the count is less than the given nth occurrence, return -1.
Else, use a loop to find the nth occurrence of the substring and return its index.

Python3

def find_nth_occurrence_1(s, sub, n):
    count = s.count(sub)
    if count < n:
        return -1
    else:
        index = -1
        for i in range(n):
            index = s.find(sub, index+1)
        return index
my_string = "hello world, how are you doing today? I hope you are doing well."
my_substring = "doing"
my_occurrence = 2
result = find_nth_occurrence_1(my_string, my_substring, my_occurrence)
print(result)

Output

53

Time Complexity: O(n)
Auxiliary Space: O(1)

METHOD 6: Using loop and counter

APPROACH:

This Approach tells how to find the index of the nth occurrence of a given substring within a given string. The approach used is a simple iteration over the string and checking whether the current substring matches the given substring. If it does, then the occurrence count is incremented. Once the count reaches the desired occurrence, the current index is returned. If no nth occurrence is found, then the function returns -1.

ALGORITHM:

1.Initialize counter as 0
2.Using a loop, search for the substring in the given string, and increase the counter every time the substring is found
3.If the counter is equal to the required occurrence, return the index of the last found substring
4.If the substring is not found enough times, return -1

Python3

def find_nth_occurrence(ini_str, substr, occurrence):
    count = 0
    for i in range(len(ini_str)):
        if ini_str[i:i+len(substr)] == substr:
            count += 1
            if count == occurrence:
                return i
    return -1

ini_str = "abababababab"
substr = "ab"
occurrence = 4

print(find_nth_occurrence(ini_str, substr, occurrence))

Output

6

Time complexity: O(n*m) where n is the length of the initial string and m is the length of the substring
Auxiliary Space: O(1)