Easy
,Math
,String
1071. Greatest Common Divisor of Strings
For two strings s
and t
, we say "t
divides s
" if and only if s = 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:
str1.length
, str2.length
<= 1000str1
and str2
consist of English uppercase letters.
class Solution {
public:
string gcdOfStrings(string str1, string str2) {
return str1 + str2 == str2 + str1 ? str1.substr(0, gcd(str1.size(), str2.size())): "";
}
};
Yen-Chi ChenWed, Feb 1, 2023
class Solution:
def gcdOfStrings(self, str1: str, str2: str) -> str:
return str1[:gcd(len(str1), len(str2))] if str1 + str2 == str2 + str1 else ""
Yen-Chi ChenWed, Feb 1, 2023
class Solution:
def gcdOfStrings(self, str1: str, str2: str) -> str:
n = min(len(str1), len(str2))
gcd = ''
while n != 0:
if len(str1) % n == 0 and len(str2) % n == 0:
gcd = str1[0:n]
print(n, len(str1), len(str2))
#check str1
flag1 = (str1 == gcd*(len(str1)//n) )
#check str2
flag2 = (str2 == gcd*(len(str2)//n) )
if flag1 and flag2:
return str1[:n]
else:
n -= 1
else:
n -= 1
return ""
玉山