Re.start() & Re.end() in Python - HackerRank Solution
Re.start() & Re.end() in Python - HackerRank Solution
Problem :
start() & end()
These expressions return the indices of the start and end of the substring matched by the group.
Code :
>>> import re >>> m = re.search(r'\d+','1234') >>> m.end() 4 >>> m.start() 0
Task :
You are given a string S.Your task is to find the indices of the start and end of string k in S.
Input Format :
The first line contains the string S.The second line contains the string k.
Constraints :
- 0 < len(s) < 100
- 0 < len(k) < len(s)
Output Format :
Print the tuple in this format: (start _index, end _index).If no match is found, print (-1, -1).
Sample Input :
aaadaa aa
Sample Output :
(0, 1) (1, 2) (4, 5)
Solution :
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | # Re.start() & Re.end() in Python - Hacker Rank Solution # Python 3 # Enter your code here. Read input from STDIN. Print output to STDOUT # Re.start() & Re.end() in Python - Hacker Rank Solution START import re S, k = input(), input() matches = re.finditer(r'(?=(' + k + '))', S) anymatch = False for match in matches: anymatch = True print((match.start(1), match.end(1) - 1)) if anymatch == False: print((-1, -1)) # Re.start() & Re.end() in Python - Hacker Rank Solution END |
Solution 2 :
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | # Re.start() & Re.end() in Python - Hacker Rank Solution # Python 3 # Enter your code here. Read input from STDIN. Print output to STDOUT # Re.start() & Re.end() in Python - Hacker Rank Solution START import re string = input() substring = input() pattern = re.compile(substring) match = pattern.search(string) if not match: print('(-1, -1)') while match: print('({0}, {1})'.format(match.start(), match.end() - 1)) match = pattern.search(string, match.start() + 1) # Re.start() & Re.end() in Python - Hacker Rank Solution END |
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