Input: str1 = "LEET", str2 = "CODE"
Output: ""

Constraints:

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

23:T5f0,Let\'s Connect on ****[LINKDIN](https://www.linkedin.com/in/mahesh-vishnoi-a4a47a193)****\n# Approach\n\n - str1+str2 == str2+str1 if and only if str1 and str2 have a gcd . \n- E.g. str1=abcabc, str2=abc, then str1+str2 = abcabcabc = str2+str1\nThis(str1+str2==str2+str1) is a requirement for the strings to **have a gcd**. If one of them is NOT a common part then gcd is "".It means we will return **empty string**\n**Proof**:-`str1 = mGCD, str2 = nGCD, str1 + str2 = (m + n)GCD = str2 + str1`\n- Both the strings are made of same substring added multiple times.\n- Since they are multiples, next step is simply to find the gcd of the lengths of 2 strings e.g. gcd(6,3) = 3, (we can use gcd function to find that)and get the substring of length 3 from either str1 or str2.\nIn c++ it will be **str1.substr(0, gcd)**\n\n![leetcode.jpg](https://assets.leetcode.com/users/images/fe51084d-a35c-433c-ab68-fcf5452067cf_1675216475.8110666.jpeg)\n# Code\n```\nclass Solution {\npublic:\n string gcdOfStrings(string str1, string str2) {\n return (str1 + str2 == str2 + str1)? \n str1.substr(0, gcd(size(str1),size(str2))): "";\n }\n};\n```\nIf you really found my solution helpful **please upvote it**, as it motivates me to post such kind of codes.\n**Let me know in comment if i can do better.**\nLet\'s Connect on **[LINKDIN](https://www.linkedin.com/in/mahesh-vishnoi-a4a47a193)**\n\n![upvote.jfif](https://assets.leetcode.com/users/images/b0bf2fa1-1680-41fc-be3f-3ba1c8745505_1675216604.7695017.jpeg)\n24:T9a0,

Greatest Common Divisor of Strings

Problem Description

Problem Description

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