# Re.start() & Re.end() in Python | HackerRank Solution

Hello coders, today we are going to solve Re.start() & Re.end() HackerRank Solution in Python.

## Objective

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``````

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 _indexend _index).
If no match is found, print `(-1, -1)`.

Sample Input

``````aaadaa
aa``````

Sample Output

``````(0, 1)
(1, 2)
(4, 5)``````

## Solution – Re.start() & Re.end() in Python

```# Enter your code here. Read input from STDIN. Print output to STDOUT
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)```

Disclaimer: The above Problem (Re.start() & Re.end()) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.