For two strings s and t, we say "t divides s" if and only if s = t + ... + t  (t concatenated with itself 1 or more times)
Given two strings str1 and str2, return the largest string x such that x divides both str1 and str2.
 
Example 1:
Input: str1 = "ABCABC", str2 = "ABC"
Output: "ABC"
Example 2:
Input: str1 = "ABABAB", str2 = "ABAB"
Output: "AB"
Example 3:
Input: str1 = "LEET", str2 = "CODE"
Output: ""
Example 4:
Input: str1 = "ABCDEF", str2 = "ABC"
Output: ""

 
Constraints:

1 <= str1.length <= 1000
1 <= str2.length <= 1000
str1 and str2 consist of English uppercase letters.

Source: leetcode | Difficulty: EASY
Tags: String algorithms

Mathematics

Starter code: class Solution:
    def gcdOfStrings(self, str1: str, str2: str) -> str:
        

Solutions:
class Solution:

	def gcdOfStrings(self, str1: str, str2: str) -> str:
		contender = ''
		for i in str1:
			contender += i
			if str1.count(contender) * len(contender) == len(str1) and str2.count(contender) * len(contender) == len(str2):
				break
		orig = contender
		ans = None
		while len(contender) <= len(str1) and len(contender) <= len(str2):
			t1 = str1.replace(contender, '')
			t2 = str2.replace(contender, '')
			if len(t1) == len(t2) == 0:
				ans = contender
			contender += orig
		return ans if ans else ''


class Solution:

	def gcdOfStrings(self, str1: str, str2: str) -> str:
		(m, n) = (len(str1), len(str2))
		(candidate, curr) = ('', '')
		for i in range(min(m, n)):
			if str1[i] != str2[i]:
				break
			else:
				curr += str1[i]
				if m % len(curr) == 0 and n % len(curr) == 0:
					if str1 == curr * (m // len(curr)) and str2 == curr * (n // len(curr)):
						candidate = curr
		return candidate


class Solution:

	def gcdOfStrings(self, str1: str, str2: str) -> str:

		def diviser(n):
			if n == 4:
				return [1, 2]
			Result = [1]
			for i in range(2, n // 2):
				if n % i == 0:
					Result.extend([int(i), int(n / i)])
			return Result

		def str_diviser(str3):
			Result = ['', str3]
			Result1 = diviser(len(str3))
			if len(Result1) > 0:
				for i in Result1:
					div = str3[:i]
					for j in range(i, len(str3), i):
						if str3[j:j + i] != div:
							break
						if j == len(str3) - i:
							Result.append(div)
			return Result
		a = str_diviser(str1)
		b = str_diviser(str2)
		if len(a) > len(b):
			(a, b) = (b, a)
		Cur = ''
		for i in a:
			if i in b and len(i) > len(Cur):
				Cur = i
		return Cur