Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/1071.html
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
andstr2
consist of English uppercase letters.
Related Topics:
String
Solution 1. Brute Force
// OJ: https://leetcode.com/problems/greatest-common-divisor-of-strings/
// Time: O(S^2)
// Space: O(S)
class Solution {
private:
bool divisible(string a, string b) {
int i = 0, j = 0, M = a.size(), N = b.size();
for (; i < M; ++i) {
if (a[i] != b[j]) return false;
if (++j == N) j = 0;
}
return true;
}
public:
string gcdOfStrings(string str1, string str2) {
string d = str1.size() < str2.size() ? str1 : str2;
for (; d.size(); d.pop_back()) {
if (str1.size() % d.size() || str2.size() % d.size()) continue;
if (divisible(str1, d) && divisible(str2, d)) return d;
}
return d;
}
};
Solution 2. GCD
// OJ: https://leetcode.com/problems/greatest-common-divisor-of-strings/
// Time: O(SH) where S is string length and H is depth of recursion
// Space: O(SH)
class Solution {
public:
string gcdOfStrings(string str1, string str2) {
if (str1.size() < str2.size()) return gcdOfStrings(str2, str1);
if (str2.empty()) return str1;
auto d = str1.substr(0, str2.size());
return d == str2 ? gcdOfStrings(str1.substr(str2.size()), str2) : "";
}
};
-
class Solution { public String gcdOfStrings(String str1, String str2) { if (str1 == null || str2 == null) return null; while (str1.length() > 0 && str2.length() > 0) { if (str1.length() > str2.length()) { String temp = str1; str1 = str2; str2 = temp; } if (str2.indexOf(str1) < 0) return ""; str2 = str2.substring(str1.length()); } return str1.length() == 0 ? str2 : str1; } } ############ class Solution { public String gcdOfStrings(String str1, String str2) { if (!(str1 + str2).equals(str2 + str1)) { return ""; } int len = gcd(str1.length(), str2.length()); return str1.substring(0, len); } private int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } }
-
// OJ: https://leetcode.com/problems/greatest-common-divisor-of-strings/ // Time: O(S^2) // Space: O(S) class Solution { private: bool divisible(string a, string b) { int i = 0, j = 0, M = a.size(), N = b.size(); for (; i < M; ++i) { if (a[i] != b[j]) return false; if (++j == N) j = 0; } return true; } public: string gcdOfStrings(string str1, string str2) { string d = str1.size() < str2.size() ? str1 : str2; for (; d.size(); d.pop_back()) { if (str1.size() % d.size() || str2.size() % d.size()) continue; if (divisible(str1, d) && divisible(str2, d)) return d; } return d; } };
-
class Solution: def gcdOfStrings(self, str1: str, str2: str) -> str: if str1 + str2 != str2 + str1: return '' n = gcd(len(str1), len(str2)) return str1[:n] ############ class Solution(object): def gcdOfStrings(self, str1, str2): """ :type str1: str :type str2: str :rtype: str """ l1, l2 = len(str1), len(str2) shorter = min(l1, l2) res = "" for i in range(shorter, 0, -1): if l1 % i or l2 % i: continue t1, t2 = l1 // i, l2 // i gcd = str1[:i] rebuild1 = gcd * t1 rebuild2 = gcd * t2 if rebuild1 == str1 and rebuild2 == str2: res = gcd break return res
-
func gcdOfStrings(str1 string, str2 string) string { if str1+str2 != str2+str1 { return "" } n := gcd(len(str1), len(str2)) return str1[:n] } func gcd(a, b int) int { if b == 0 { return a } return gcd(b, a%b) }