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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 and str2 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)
}