Question

1071. Greatest Common Divisor of Strings

中文文档

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: ""

 

Constraints:

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

Solutions

Solution 1

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 ''

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);
  }
}

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);
  }
};

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)
}

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()
  }
}

Solution 2

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]