Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:


Input: haystack = "hello", needle = "ll"
Output: 2


Example 2:


Input: haystack = "aaaaa", needle = "bba"
Output: -1


Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().

Difficulty: EASY
Starter code: class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        

Solutions:
class Solution:

	def strStr(self, haystack, needle):
		if haystack == '' and needle == '':
			return 0
		if needle == '':
			return 0
		if haystack == '' or needle == '' or len(haystack.split(needle)) == 1:
			return -1
		return len(haystack.split(needle)[0])


class Solution:

	def strStr(self, haystack, needle):
		return haystack.find(needle)


class Solution:

	def strStr(self, haystack, needle):
		for i in range(len(haystack) - len(needle) + 1):
			if haystack[i:i + len(needle)] == needle:
				return i
		return -1


class Solution:

	def strStr(self, haystack, needle):
		if not needle:
			return 0
		for i in range(0, len(haystack) - len(needle) + 1):
			match = True
			for j in range(len(needle)):
				if haystack[i + j] != needle[j]:
					match = False
					break
			if match:
				return i
		return -1


class Solution:

	def strStr(self, haystack, needle):
		if needle == '':
			return 0
		for i in range(len(haystack) - len(needle) + 1):
			for j in range(len(needle)):
				if haystack[i + j] == needle[j]:
					if j == len(needle) - 1:
						return i
				else:
					break
		return -1


class Solution:

	def makeNext(self, s):
		k = -1
		i = 0
		next = [None for i in range(len(s))]
		next[0] = -1
		while i < len(s) - 1:
			while k >= 0 and s[i] != s[k]:
				k = next[k]
			i += 1
			k += 1
			if s[i] == s[k]:
				next[i] = next[k]
			else:
				next[i] = k
		return next

	def strStr(self, haystack, needle):
		if needle == '':
			return 0
		Next = self.makeNext(needle)
		length = len(needle)
		for i in range(len(haystack)):
			count = 0
			while haystack[i + count] == needle[count]:
				count += 1
				if count + i >= len(haystack) and count < length:
					return -1
					break
				if count >= length:
					return i
					break
		return -1


class Solution:

	def strStr(self, haystack, needle):
		if not needle:
			return 0
		nexts = [0] * len(needle)
		nexts[0] = -1
		if len(needle) > 1:
			nexts[1] = 0
		left = 0
		right = 2
		while right < len(needle):
			if needle[left] == needle[right - 1]:
				nexts[right] = left + 1
				left += 1
				right += 1
			elif nexts[left] > 0:
				left = nexts[left]
			else:
				nexts[right] = 0
				left = 0
				right += 1
		i = 0
		j = 0
		while i <= len(haystack) - len(needle):
			if j == len(needle):
				return i
			if haystack[i + j] == needle[j]:
				j += 1
			elif nexts[j] > 0:
				i = i + j - nexts[j]
				j = nexts[j]
			else:
				i = i + 1
				j = 0
		return -1


class Solution:

	def myNext(self, T):
		next = [-1]
		l = len(T)
		(i, j) = (0, -1)
		while i < l:
			if j == -1 or T[i] == T[j]:
				i += 1
				j += 1
				next.append(int(j))
			else:
				j = next[j]
		return next

	def strStr(self, haystack, needle):
		if needle == '':
			return 0
		arr = Solution().myNext(needle)
		l1 = len(haystack)
		l2 = len(needle)
		(i, j) = (0, 0)
		while i < l1 and j < l2:
			if j == -1 or haystack[i] == needle[j]:
				i += 1
				j += 1
				if j == l2:
					return i - l2
			else:
				j = arr[j]
		return -1


class Solution(object):

	def strStr(self, haystack, needle):
		if len(haystack) == len(needle):
			if haystack == needle:
				return 0
			else:
				return -1
		for i in range(0, len(haystack)):
			k = i
			j = 0
			while j < len(needle) and k < len(haystack) and (haystack[k] == needle[j]):
				j += 1
				k += 1
			if j == len(needle):
				return i
		return -1 if needle else 0


class Solution:

	def strStr(self, haystack, needle):
		if len(haystack) < len(needle):
			return -1
		if needle == '':
			return 0
		for i in range(len(haystack)):
			if haystack[i] == needle[0]:
				res = True
				j = 1
				try:
					while j < len(needle):
						if haystack[i + j] == needle[j]:
							j += 1
						else:
							res = False
							break
					if res:
						return i
				except:
					return -1
		return -1


class Solution:

	def strStr(self, haystack, needle):
		(i, j, m, n) = (-1, 0, len(haystack), len(needle))
		jump = [-1] * n
		while j < n - 1:
			if i == -1 or needle[i] == needle[j]:
				(i, j) = (i + 1, j + 1)
				jump[j] = jump[i] if needle[i] == needle[j] else i
			else:
				i = jump[i]
		(i, j) = (0, 0)
		while i < m and j < n:
			if j == -1 or haystack[i] == needle[j]:
				(i, j) = (i + 1, j + 1)
			else:
				j = jump[j]
		return i - j if j == n else -1


class Solution:

	def strStr(self, haystack, needle):
		next = self.next(needle)
		i = 0
		j = 0
		while i < len(haystack) and j < len(needle):
			if j == -1 or haystack[i] == needle[j]:
				i += 1
				j += 1
			else:
				j = next[j]
		if j == len(needle):
			return i - j
		else:
			return -1

	def next(self, p):
		next = []
		next.append(-1)
		i = 0
		k = -1
		while i < len(p) - 1:
			if k == -1 or p[i] == p[k]:
				i += 1
				k += 1
				next.append(k)
			else:
				k = next[k]
		return next


class Solution:

	def strStr(self, haystack, needle):
		if not needle:
			return 0
		if not haystack:
			return -1
		if len(haystack) < len(needle):
			return -1
		N = len(haystack)
		for i in range(N):
			if needle == haystack[i:i + len(needle)]:
				return i
		return -1


class Solution:

	def strStr(self, haystack, needle):
		if needle == '':
			return 0
		substr_len = len(needle)
		for (index, char) in enumerate(haystack):
			if haystack[index:index + substr_len] == needle:
				return index
		return -1


class Solution:

	def prefixTable(self, s):
		t = [0 for _ in s]
		j = 0
		i = 1
		while i < len(s):
			if j == 0 and s[i] != s[j]:
				t[i] = 0
				i += 1
			elif s[i] == s[j]:
				t[i] = j + 1
				i += 1
				j += 1
			elif j > 0 and s[j] != s[i]:
				j = t[j - 1]
		return t

	def find(self, text, pattern):
		ans = []
		t = self.prefixTable(pattern)
		i = j = 0
		while i < len(text):
			if text[i] == pattern[j]:
				i += 1
				j += 1
			if j == len(pattern):
				ans.append(i - len(pattern))
				j = t[j - 1]
			elif i < len(text) and text[i] != pattern[j]:
				if j > 0:
					j = t[j - 1]
				else:
					i += 1
		return ans

	def strStr(self, haystack, needle):
		if len(needle) == 0:
			return 0
		ans = self.find(haystack, needle)
		return ans[0] if ans else -1