LeetCode 1071. Greatest Common Divisor of Strings Solution in Java, C++, Python & More | Explanation + Code

Description

For two strings s and t, we say "t divides s" if and only if s = t + t + t + ... + 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: ""

Example 4:

Input: str1 = "AAAAAB", str2 = "AAA"

Output: ""​​​​​​​

 

Constraints:

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

Solutions

Solution 1

PythonJavaC++GoRust

class Solution: def gcdOfStrings(self, str1: str, str2: str) -> str: def check(a, b): c = "" while len(c) < len(b): c += a return c == b for i in range(min(len(str1), len(str2)), 0, -1): t = str1[:i] if check(t, str1) and check(t, str2): return t return ''(code-box)

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); } }(code-box)

class Solution { public: string gcdOfStrings(string str1, string str2) { if (str1 + str2 != str2 + str1) return ""; int n = __gcd(str1.size(), str2.size()); return str1.substr(0, n); } };(code-box)

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) }(code-box)

impl Solution { pub fn gcd_of_strings(str1: String, str2: String) -> String { if str1.clone() + &str2 != str2.clone() + &str1 { return String::from(""); } fn gcd(a: usize, b: usize) -> usize { if b == 0 { return a; } gcd(b, a % b) } let (m, n) = (str1.len().max(str2.len()), str1.len().min(str2.len())); str1[..gcd(m, n)].to_string() } }(code-box)

Solution 2

Python

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](code-box)