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Find all the patterns of “1(0+)1” in a given string using Python Regex
• Last Updated : 29 Dec, 2020

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.
 `# 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 ``    ``# untill 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
```

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