# 1071. Greatest Common Divisor of Strings

## Description

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:

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

## Solutions

• class Solution {
public String gcdOfStrings(String str1, String str2) {
if (!(str1 + str2).equals(str2 + str1)) {
return "";
}
int len = gcd(str1.length(), str2.length());
return str1.substring(0, len);
}

private int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
}

• class Solution {
public:
string gcdOfStrings(string str1, string str2) {
if (str1 + str2 != str2 + str1) return "";
int n = __gcd(str1.size(), str2.size());
return str1.substr(0, n);
}
};

• class Solution:
def gcdOfStrings(self, str1: str, str2: str) -> str:
def check(a, b):
c = ""
while len(c) < len(b):
c += a
return c == b

for i in range(min(len(str1), len(str2)), 0, -1):
t = str1[:i]
if check(t, str1) and check(t, str2):
return t
return ''


• func gcdOfStrings(str1 string, str2 string) string {
if str1+str2 != str2+str1 {
return ""
}
n := gcd(len(str1), len(str2))
return str1[:n]
}

func gcd(a, b int) int {
if b == 0 {
return a
}
return gcd(b, a%b)
}

• impl Solution {
pub fn gcd_of_strings(str1: String, str2: String) -> String {
if str1.clone() + &str2 != str2.clone() + &str1 {
return String::from("");
}
fn gcd(a: usize, b: usize) -> usize {
if b == 0 {
return a;
}
gcd(b, a % b)
}

let (m, n) = (str1.len().max(str2.len()), str1.len().min(str2.len()));
str1[..gcd(m, n)].to_string()
}
}