88 lines
2.1 KiB
C++

// Source : https://leetcode.com/problems/greatest-common-divisor-of-strings/
// Author : Hao Chen
// Date : 2020-07-19
/*****************************************************************************************************
*
* For strings S and T, we say "T divides S" if and only if S = T + ... + T (T concatenated with
* itself 1 or more times)
*
* Return the largest string X such that X divides str1 and X divides 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: ""
*
* Note:
*
* 1 <= str1.length <= 1000
* 1 <= str2.length <= 1000
* str1[i] and str2[i] are English uppercase letters.
******************************************************************************************************/
class Solution {
private:
// Euclidean algorithm
// https://en.wikipedia.org/wiki/Euclidean_algorithm
// recursive way
int findGCD_r(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
// non-recursive way
int findGCD(int a, int b) {
int t = 1;
while(t != 0) {
t = a % b;
a = b;
b = t;
}
return a;
}
bool isStrRepeatByLen(string& s, int len) {
if (s.size() == len) return true;
if (s.size() % len != 0 ) return false;
for (int l=0; l<len; l++) {
for (int i=1; i<s.size()/len; i++) {
if (s[l] != s[i*len+l]) return false;
}
}
return true;
}
bool strPrefixComp(string& s1, string &s2, int len){
for(int i=0; i<len; i++) {
if (s1[i] != s2[i]) return false;
}
return true;
}
public:
string gcdOfStrings(string s1, string s2) {
int gcd = findGCD(s1.size(), s2.size());
if (strPrefixComp(s1, s2, gcd) &&
isStrRepeatByLen(s1, gcd)&&
isStrRepeatByLen(s2, gcd) ){
return s2.substr(0,gcd);
}
return "";
}
};