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` 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.

1071. Greatest Common Divisor of Strings
``````struct Solution;

impl Solution {
fn gcd_of_strings(str1: String, str2: String) -> String {
let n1 = str1.len();
let n2 = str2.len();
let mut n = usize::min(n1, n2);
while n > 0 {
if n1 % n != 0 || n2 % n != 0 {
n -= 1;
continue;
}
let s1 = &str1[0..n];
let s2 = &str2[0..n];
if s1 != s2 {
n -= 1;
continue;
}
let k1 = n1 / n;
let k2 = n2 / n;
if str1.matches(s1).count() == k1 && str2.matches(s2).count() == k2 {
return s1.to_string();
}
n -= 1;
}
"".to_string()
}
}

#[test]
fn test() {
let str1 = "ABCABC".to_string();
let str2 = "ABC".to_string();
let res = "ABC".to_string();
assert_eq!(Solution::gcd_of_strings(str1, str2), res);
let str1 = "ABABAB".to_string();
let str2 = "ABAB".to_string();
let res = "AB".to_string();
assert_eq!(Solution::gcd_of_strings(str1, str2), res);
let str1 = "LEET".to_string();
let str2 = "CODE".to_string();
let res = "".to_string();
assert_eq!(Solution::gcd_of_strings(str1, str2), res);
}
``````