Comment on page
1071. Greatest Common Divisor of Strings
Easy
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 <= 1000str1andstr2consist of English uppercase letters.
func gcdOfStrings(str1 string, str2 string) string {
l := gcd(len(str1), len(str2))
if str1[:l] != str2[:l] {
return ""
}
for i:=0; i<=len(str1)-l; i+=l {
if str1[i:i+l] != str1[:l] {
return ""
}
}
for i:=0; i<=len(str2)-l; i+=l {
if str2[i:i+l] != str2[:l] {
return ""
}
}
return str1[:l]
}
func gcd(a, b int) int {
if b == 0 { return a }
return gcd(b, a % b)
}