In this Kata, you will count the number of times the first string occurs in the second. 

```Haskell
solve("zaz","zazapulz") = 4 because they are ZAZapulz, ZAzapulZ, ZazApulZ, zaZApulZ
```

More examples in test cases. 

Good luck!

Please also try [Simple time difference](https://www.codewars.com/kata/5b76a34ff71e5de9db0000f2)

Difficulty: EASY
Starter code: def solve(a, b):
	

Solutions:
def solve(a, b):
	if len(a) is 1:
		return b.count(a)
	index = [x for x in range(len(b) - 1) if b[x] is a[0]]
	return sum((solve(a[1:], b[x + 1:]) for x in index))


def solve(a, b, i=0, j=0):
	return i == len(a) or sum((solve(a, b, i + 1, x + 1) for x in range(j, len(b) - (len(a) - i - 1)) if a[i] == b[x]))


from functools import lru_cache

def solve(a, b):
	(l_a, l_b, rec) = (len(a), len(b), lru_cache(maxsize=None)(lambda x, y: x == l_a or sum((rec(x + 1, i + 1) for i in range(y, l_b) if b[i] == a[x]))))
	return rec(0, 0)


def solve(a, b):
	if not a:
		return 1
	if not b:
		return 0
	if b[0].lower() == a[0].lower():
		return solve(a, b[1:]) + solve(a[1:], b[1:])
	else:
		return solve(a, b[1:])


def solve(a, b):
	if a[0] in b:
		if len(a) > 1:
			return solve(a[1:], b[b.index(a[0]) + 1:]) + solve(a, b[b.index(a[0]) + 1:])
		else:
			print(b.count(a))
			return b.count(a)
	else:
		return 0


def solve(a, b, i=0, j=0):
	return 1 if i == len(a) else 0 if j == len(b) else solve(a, b, i, j + 1) if a[i] != b[j] else solve(a, b, i + 1, j + 1) + solve(a, b, i, j + 1)


def solve_dp(needle, haystack):
	(N, H) = (len(needle), len(haystack))
	prefixes = [[1] + N * [0] for h in range(H)]

	def match(h, n):
		return haystack[h] == needle[n - 1]
	prefixes[0][1] = 1 if match(0, 1) else 0
	for h in range(1, H):
		max_prefix_len = min(h + 1, N)
		for n in range(1, max_prefix_len + 1):
			with_current = prefixes[h - 1][n - 1] if match(h, n) else 0
			without_current = prefixes[h - 1][n]
			prefixes[h][n] = with_current + without_current
	return prefixes[H - 1][N]

def solve_recursive(needle, haystack):
	if not needle or needle == haystack:
		return 1
	if not haystack:
		return 0
	(needle_head, *needle_tail) = needle
	(haystack_head, *haystack_tail) = haystack
	match = needle_head == haystack_head
	head_matches = solve_recursive(needle_tail, haystack_tail) if match else 0
	tail_matches = solve_recursive(needle, haystack_tail)
	return head_matches + tail_matches

def solve(needle, haystack, method=solve_dp):
	return method(needle, haystack)


from itertools import combinations as c

def solve(a, b):
	tot = 0
	a = tuple(a)
	b = [i for i in b if i in a]
	for k in range(len(b)):
		if b[k] != a[0]:
			continue
		for kk in range(k + 1, len(b)):
			if b[kk] != a[1]:
				continue
			for i in c(b[kk + 1:], len(a) - 2):
				if i == a[2:]:
					tot += 1
	return tot


def solve(a, b):
	if len(a) == 1:
		return b.count(a)
	r = 0
	for i in range(len(b) - len(a) + 1):
		if b[i] == a[0]:
			r += solve(a[1:], b[i + 1:])
	return r


def solve(a, b):
	if not a:
		return 1
	(x, c) = (a[0], 0)
	for i in range(len(b)):
		if b[i] == x:
			c += solve(a[1:], b[i + 1:])
	return c