LeetCode 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 (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.

Solution

from math import gcd
class Solution:
    def gcdOfStrings(self, str1: str, str2: str) -> str:
        if str1 + str2 != str2 + str1:
            return ""
        elif str1 == str2:
            return str1
        else:
            length_by_gcd = gcd(len(str1), len(str2))
            return str1[:length_by_gcd]

Explanation:

Example explanation:

input: “ABCABC”, “ABC”

gcdOfString( “ABCABC”, “ABC” )　　　　　# gcd( 6, 3 ) = 3

-> gcdOfString( “ABC”, “ABC” )

-> match, return “ABC” as greatest common divisor of string

input: “ABABAB”, “AB”

gcdOfString( “ABABAB, “AB” ) 　　　　　# gcd( 6, 2 ) = 2

-> gcdOfString( “AB”, “AB” )

-> match, return “AB” as greatest common divisor of string

input: “LEET”, “CODE”

gcdOfString( “LEET, “CODE” ) 　　　　　　# “LEETCODE” =/= “CODELEET”

-> reject, “LEET” and “CODE” have no common factor string.