【朱迪的LeetCode刷题笔记】】1071. Greatest Common Divisor of Strings #Python ...

Judie

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

Judy
Python stupid way calculating gcd

class Solution(object):
    def gcdOfStrings(self, str1, str2):
        """
        :type str1: str
        :type str2: str
        :rtype: str
        """
        l1 = len(str1)
        l2 = len(str2)
        l0 = min(l1, l2)
        r = ""
        for i0 in range(1,l0+1):
            if l1 % i0 == 0 and l2 % i0 == 0 and str1[:i0] == str2[:i0] and str1.count(str1[:i0]) == l1 / i0 and str2.count(str2[:i0]) == l2 / i0:
                r = str2[:i0]
        return r

Sol1
Python amazing approach!
https://leetcode.com/problems/gr ... p;envId=leetcode-75

class Solution(object):
    def gcdOfStrings(self, str1, str2):
        """
        :type str1: str
        :type str2: str
        :rtype: str
        """

        def gcd(int1, int2):
            if (int2 == 0):
                return int1
            else:
                return gcd(int2, int1 % int2)

        if str1 + str2 != str2 + str1:
            return ""
        maxlen = gcd(len(str1), len(str2))
        return str1[:maxlen]

Gray
C++ smarter way calculating gcd

class Solution {
public:
int gcd(int a, int b) // The function runs recursive in nature to return GCD
{

    if (a == 0) // If a becomes zero
       return b; // b is the GCD
    if (b == 0)// If b becomes zero
       return a;// a is the GCD

    if (a == b) // The case of equal numbers
        return a; // return any one of them

   // Apply case of substraction
    if (a > b) // if a is greater subtract b
        return gcd(a-b, b);
    return gcd(a, b-a); //otherwise subtract a
}
    string gcdOfStrings(string str1, string str2) {
        int c = gcd(str1.size(),str2.size());
        bool test = true;
        string temp = str2.substr(0,c);
        for(int i = 0; i < str1.size()/c; i++){
            if(str1.substr(ic,c) != temp){
                return "";
            }
        }
        for(int i = 0; i < str2.size()/c; i++){
            if(str2.substr(ic,c) != temp){
                return "";
            }
        }
        return temp;
    }
};

Lemma & Proof