Regex | Find Yellorn

Source: https://yellorn.com/programming/regex-find-yellorn

At Yellorn, we always want to find out how popular we are getting every day and have scraped conversations from popular sites. Each conversation fits in 1 line and there are N such conversations. Each conversation has at most 1 word that says yellorn (all in lowercase). We would like you to help us figure out whether a conversation:

1. Starts with yellorn
2. Ends with yellorn
3. Start and ends with yellorn

Input Format

First line of the input contains an integer, N. Then N lines follow.
From the second line onwards, each line contains a set of W words separated by a single space

Constraints

• 1 <= N <= 10
• 1 <= W <= 100
• All the characters in W are lowercase alphabet characters.
• If C is the count of the characters in W, then 1 <= C <= 20

Output Format

For every line,

1. Print 1 if the conversation starts with yellorn
2. Print 2 if the conversation ends with yellorn
3. Print 0 if the conversation starts and ends with yellorn
4. Print -1 if none of the above.

Explanation

The first conversation ends with yellorn and hence 2
The second conversation starts with yellorn and hence 1
The third conversation has only one word, it starts and ends with yellorn and hence 0.
The fourth conversation satisfies none of the above properties and hence -1.

Testcase

Input

4
i love hackerrank
hackerrank is an awesome place for programmers
hackerrank
i think hackerrank is a great place to hangout

Output

2
1
0
-1

Solution

import sys
import re

first_pattern = r"^yellorn"
last_pattern = r"yellorn$"
for i in range(int(input())):
    text = sys.stdin.readline()
    first = re.search(first_pattern, text)
    last = re.search(last_pattern, text)
    result = -1
    if first and last:
        result = 0
    elif first:
        result = 1
    elif last:
        result = 2
    print(result)