Description
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
and str2
consist of English uppercase letters.
Solutions
Solution 1
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13 | 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 ''
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13 | 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);
}
}
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| 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);
}
};
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14 | 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)
}
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16 | 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()
}
}
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Solution 2