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 :
- Search a first sub-string in original string which follows ’10+1′ pattern using re.search(regex,string) method.
- 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.
- 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) |
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Output:
2
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