Problem
For two strings s
and t
, we say “t
divides s
” if and only if s = 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
.
Examples
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: ""
Solution
Method 1 - Run the gcd
To solve the problem of finding the largest string x
such that x
divides both str1
and str2
, we can employ the greatest common divisor (GCD) concept:
- Check Divisibility:
- We need to check if a substring
x
can repeatedly form bothstr1
andstr2
.
- We need to check if a substring
- GCD of Lengths:
- The length of the largest
x
that can divide bothstr1
andstr2
will be the greatest common divisor (GCD) of their lengths. This follows from the property of GCD in number theory.
- The length of the largest
- Construct and Verify:
- Construct the substring
x
using the firstGCD(len(str1), len(str2))
characters of either string. - Verify if this substring can repeatedly form both
str1
andstr2
.
- Construct the substring
Code
Java
public class Solution {
public String gcdOfStrings(String str1, String str2) {
if (!(str1 + str2).equals(str2 + str1)) {
return "";
}
int gcdLength = gcd(str1.length(), str2.length());
return str1.substring(0, gcdLength);
}
private int gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
}
Python
class Solution:
def gcdOfStrings(self, str1: str, str2: str) -> str:
if not (str1 + str2 == str2 + str1):
return ""
def gcd(a: int, b: int) -> int:
while b:
a, b = b, a % b
return a
gcd_length = gcd(len(str1), len(str2))
return str1[:gcd_length]
Complexity
- ⏰ Time complexity:
O(n + m + gcd(n, m))
, wheren
andm
are the lengths ofstr1
andstr2
respectively. This accounts for computing the GCD and verifying the constructed substring. - 🧺 Space complexity:
O(gcd(n, m))
, for storing the resulting substring.