Hackerrank - Find A Sub-Word solution

We define a word character to be any of the following:

• An English alphabetic letter (i.e., a-z and A-Z).
• A decimal digit (i.e., 0-9).
• An underscore (i.e., _, which corresponds to ASCII value ).

We define a word to be a contiguous sequence of one or more word characters that is preceded and succeeded by one or more occurrences of non-word-characters or line terminators. For example, in the string I l0ve-cheese_?, the words are I, l0ve, and cheese_.

We define a sub-word as follows:

• A sequence of word characters (i.e., English alphabetic letters, digits, and/or underscores) that occur in the same exact order (i.e., as a contiguous sequence) inside another word.
• It is preceded and succeeded by word characters only.

Given  sentences consisting of one or more words separated by non-word characters, process  queries where each query consists of a single string, . To process each query, count the number of occurrences of  as a sub-word in all  sentences, then print the number of occurrences on a new line.

Input Format

The first line contains an integer, n, denoting the number of sentences.
Each of the  subsequent lines contains a sentence consisting of words separated by non-word characters.
The next line contains an integer, , denoting the number of queries.
Each line  of the  subsequent lines contains a string, , to check.

Constraints

Output Format

For each query string, , print the total number of times it occurs as a sub-word within all words in all  sentences.

Sample Input

1
existing pessimist optimist this is
1
is

Sample Output

3

Explanation

We must count the number of times  is occurs as a sub-word in our  input sentence(s):

• occurs  time as a sub-word of existing.
• occurs  time as a sub-word of pessimist.
• occurs  time as a sub-word of optimist.
• While  is a substring of the word this, it's followed by a blank space; because a blank space is non-alphabetic, non-numeric, and not an underscore, we do not count it as a sub-word occurrence.
• While  is a substring of the word is in the sentence, we do not count it as a match because it is preceded and succeeded by non-word characters (i.e., blank spaces) in the sentence. This means it doesn't count as a sub-word occurrence.

Next, we sum the occurrences of  as a sub-word of all our words as . Thus, we print  on a new line.

Solution in Python

import re

sentences = []
words = []

no_of_sentences = int(input())

for i in range(no_of_sentences):
    sentences.append(input())

no_of_words = int(input())

for i in range(no_of_words):
    words.append(input())

for word in words:
    print(len(re.findall("[\w]+%s[\w]+"%word, "\n".join(sentences))))