Find all the patterns of “1(0+)1” in a given string using Python Regex

A string contains patterns of the form 1(0+)1 where (0+) represents any non-empty consecutive sequence of 0’s. Count all such patterns. The patterns are allowed to overlap. Note : It contains digits and lowercase characters only. The string is not necessarily a binary. 100201 is not a valid pattern. Examples:

Input : 1101001
Output : 2

Input : 100001abc101
Output : 2

We have existing solution for this problem please refer Find all the patterns of “1(0+)1” in a given string link. Another set containing similar solution using regex in java is also published. We will solve this problem quickly in python using Regex. Approach is very simple :

1. Search a first sub-string in original string which follows ’10+1′ pattern using re.search(regex,string) method.
2. substr = re.search(regex,string) return None if it doesn’t find given regex as sub-string in original string otherwise it returns first matched sub-string which follows ’10+1′ pattern. substr.start() gives us starting index of matched regex and substr.end() gives us ending index of matched regex.
3. Whenever we find regex as sub-string then increase count by 1 and again search for given regex starting from ending index of previous sub-string.

Python3

# Python program to Find all the patterns
# of “1(0+)1” in a given string using Python Regex

import re

# Function to Find all the patterns
# of “1(0+)1” in a given string
def extract(input):

    # search regex '10+1' in original string
    # search() function return first occurrence
    # of regex '10+1' otherwise None
    # '10+1' means sub-string starting and ending with 1
    # and atleast 1 or more zeros in between
    count=0
    substr = re.search('10+1',input)

    # search for regex in original string
    # until we are done with complete string
    while substr!=None:
        # if we find any occurrence then increase count by 1
        count=count+1

        # find next occurrence just after previous
        # sub-string
        # for first occurrence 101, substr.start()=1
        # substr.end()=4
        input = input[(substr.end()-1):]
        substr = re.search('10+1',input)
    print (count)

# Driver program
if __name__ == "__main__":
    input = '1101001'
    extract(input)

Output:

2

Time Complexity : O(N)

Space Complexity : O(1)