Complete the solution so that it returns the number of times the search_text is found within the full_text.

```python
search_substr( fullText, searchText, allowOverlap = true )
```

so that overlapping solutions are (not) counted. If the searchText is empty, it should return `0`. Usage examples:

```python
search_substr('aa_bb_cc_dd_bb_e', 'bb') # should return 2 since bb shows up twice
search_substr('aaabbbcccc', 'bbb') # should return 1
search_substr( 'aaa', 'aa' ) # should return 2
search_substr( 'aaa', '' ) # should return 0
search_substr( 'aaa', 'aa', false ) # should return 1
```

Source: codewars | Difficulty: EASY
Tags: String algorithms

Starter code: def search_substr(full_text, search_text, allow_overlap=True):
	

Solutions:
import re

def search_substr(full_text, search_text, allow_overlap=True):
	if not full_text or not search_text:
		return 0
	return len(re.findall(f'(?=({search_text}))' if allow_overlap else search_text, full_text))


import re

def search_substr(origin, target, overlap=True):
	if not target:
		return 0
	(start, end) = map(re.escape, (target[0], target[1:]))
	patternStr = '{}{}'.format(start, f'(?={end})' if overlap else end)
	return sum((1 for _ in re.finditer(patternStr, origin)))


def search_substr(f_t, s_t, allow_overlap=True):
	if not s_t or not f_t:
		return 0
	if allow_overlap:
		return sum((1 for i in range(len(f_t) - len(s_t) + 1) if f_t[i:i + len(s_t)] == s_t))
	return f_t.count(s_t)


import re

def search_substr(full_text, search_text, allow_overlap=True):
	if allow_overlap:
		try:
			return len(re.findall(search_text[0] + '(?={})'.format(search_text[1:]), full_text))
		except IndexError:
			return 0
	else:
		return len(re.findall(search_text, full_text)) if search_text else 0


import re

def search_substr(full_text, search_text, allow_overlap=True):
	if not full_text or not search_text:
		return 0
	if len(search_text) == 1 or not allow_overlap:
		pattern = re.compile(search_text)
	else:
		pattern = re.compile(f'(?<={search_text[:-1]}){search_text[-1]}')
	return len(list(re.findall(pattern, full_text)))


def search_substr(ft, st, allow_overlap=True):
	if not st:
		return 0
	(i, L, M, r, x) = (0, len(ft), len(st), 0, 1)
	if not allow_overlap:
		x = M
	while i + M <= L:
		if ft[i:i + M] == st:
			(r, i) = (r + 1, i + x - 1)
		i += 1
	return r


def search_substr(full_text, search_text, allow_overlap=True):
	if search_text == '':
		return 0
	if allow_overlap:
		return len([1 for i in range(len(full_text)) if full_text.startswith(search_text, i)])
	return full_text.count(search_text)


def search_substr(s, sub, overlap=True):
	return 0 if not s or not sub else sum((s[i:].startswith(sub) for i in range(len(s)))) if overlap else len(s.split(sub)) - 1