Write a method that will search an array of strings for all strings that contain another string, ignoring capitalization. Then return an array of the found strings. 

The method takes two parameters, the query string and the array of strings to search, and returns an array. 

If the string isn't contained in any of the strings in the array, the method returns an array containing a single string: "Empty" (or `Nothing` in Haskell, or "None" in Python and C)

### Examples
If the string to search for is "me", and the array to search is ["home", "milk", "Mercury", "fish"], the method should return ["home", "Mercury"].

Source: codewars | Difficulty: EASY
Tags: String algorithms

Fundamentals

Data structures

Starter code: def word_search(query, seq):
	

Solutions:
def word_search(query, seq):
	return [x for x in seq if query.lower() in x.lower()] or ['None']


def word_search(query, seq):
	query = query.lower()
	result = [x for x in seq if query in x.lower()]
	return result if result else ['None']


import re

def word_search(query, seq):
	return [w for w in seq if re.search(query, w, re.I)] or ['None']


def word_search(query, seq):
	l = [i for i in seq if query.lower() in i.lower()]
	return [l, ['None']][not l]


def word_search(query, seq):
	query = query.lower()
	return [word for word in seq if query in word.lower()] or ['None']


def word_search(q, l):
	return [w for w in l if q.lower() in w.lower()] or ['None']


def word_search(query, seq):
	query = query.lower()
	return [a for a in seq if query in a.lower()] or ['None']


def word_search(query, seq):
	return [i for i in seq if query.lower() in i.lower()] or ['None']


def word_search(query, seq):
	array = []
	for x in seq:
		if query.lower() in x.lower():
			array.append(x)
		else:
			pass
	if array != []:
		return array
	else:
		array = ['None']
		return array