1071. Greatest Common Divisor of Strings
For two strings s
and t
, we say "t
divides s
" if and only if s = t + t + t + ... + t + t
(i.e., t
is concatenated with itself one 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: ""
Constraints:
1 <= str1.length, str2.length <= 1000
str1
andstr2
consist of English uppercase letters.
Solution:
class Solution:
def gcdOfStrings(self, str1: str, str2: str) -> str:
len1, len2 = len(str1), len(str2)
def isDivisor(l):
if len1 % l or len2 % l:
return False
f1, f2 = len1 // l , len2 // l
return str1[:l]*f1 == str1 and str1[:l]*f2 == str2
for l in range(min(len1, len2), 0, -1):
if isDivisor(l):
return str1[:l]
return ""