Problem Statement:
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
andstr2
consist of English uppercase letters.
Understanding the Problem:
To solve this problem, we need to find the largest string that, when repeated, can form both str1
and str2
. This problem is akin to finding the greatest common divisor (GCD) of two numbers, but instead, we are dealing with strings.
Initial Thought Process:
We need to determine if the two strings have a common pattern that can repeatedly form both strings. If str1 + str2
is equal to str2 + str1
, then there exists a common divisor string. The length of this common string would be the GCD of the lengths of str1
and str2
.
Basic Solution:
The basic approach involves concatenating the strings and checking for the condition str1 + str2 === str2 + str1
. If this holds true, the solution would be the substring of str1
up to the length of the GCD of their lengths.
Code:
function gcdOfStringsBasic(str1: string, str2: string): string { // Helper function to find the greatest common divisor of two numbers function gcd(a: number, b: number): number { if (b === 0) { return a; } return gcd(b, a % b); } // Check if str1 + str2 is the same as str2 + str1 if (str1 + str2 !== str2 + str1) { return ""; } // Find the greatest common divisor of the lengths of str1 and str2 let gcdLength = gcd(str1.length, str2.length); // Return the substring of str1 or str2 from 0 to gcdLength return str1.substring(0, gcdLength);
}
Time Complexity Analysis:
-
Time Complexity: O(n + m), where n is the length of
str1
and m is the length ofstr2
. The concatenation check takes O(n + m) time, and the GCD computation takes O(log(min(n, m))). - Space Complexity: O(n + m) for the concatenated strings and O(1) for the GCD computation.
Limitations:
The basic solution is efficient given the problem constraints. It leverages string concatenation and GCD computation to achieve the desired result.
Optimized Solution:
The basic solution is already quite optimal, but we can ensure that the code is as efficient and clean as possible. We will reuse the GCD computation function and include a more streamlined approach to check for the common divisor.
Code:
function gcdOfStringsOptimized(str1: string, str2: string): string { // Helper function to find the greatest common divisor of two numbers function gcd(a: number, b: number): number { while (b !== 0) { [a, b] = [b, a % b]; } return a; } // Check if str1 + str2 is the same as str2 + str1 if (str1 + str2 !== str2 + str1) { return ""; } // Find the greatest common divisor of the lengths of str1 and str2 let gcdLength = gcd(str1.length, str2.length); // Return the substring of str1 or str2 from 0 to gcdLength return str1.substring(0, gcdLength);
}
Time Complexity Analysis:
-
Time Complexity: O(n + m), where n is the length of
str1
and m is the length ofstr2
. The concatenation check takes O(n + m) time, and the GCD computation takes O(log(min(n, m))). - Space Complexity: O(1) for the GCD computation, as the concatenation check does not store additional strings.
Improvements Over Basic Solution:
- The optimized solution uses an iterative approach for the GCD function, which can be more efficient than the recursive approach in some environments.
- It also avoids creating new strings unnecessarily, making it slightly more space efficient.
Edge Cases and Testing:
Edge Cases:
-
str1
is a multiple ofstr2
. -
str2
is a multiple ofstr1
. -
str1
andstr2
have no common divisor. -
str1
orstr2
is empty.
Test Cases:
console.log(gcdOfStringsBasic("ABCABC", "ABC")); // "ABC"
console.log(gcdOfStringsBasic("ABABAB", "ABAB")); // "AB"
console.log(gcdOfStringsBasic("LEET", "CODE")); // ""
console.log(gcdOfStringsBasic("ABCDEF", "ABC")); // ""
console.log(gcdOfStringsBasic("AAAAAA", "AA")); // "AA"
console.log(gcdOfStringsBasic("AA", "A")); // "A" console.log(gcdOfStringsOptimized("ABCABC", "ABC")); // "ABC"
console.log(gcdOfStringsOptimized("ABABAB", "ABAB")); // "AB"
console.log(gcdOfStringsOptimized("LEET", "CODE")); // ""
console.log(gcdOfStringsOptimized("ABCDEF", "ABC")); // ""
console.log(gcdOfStringsOptimized("AAAAAA", "AA")); // "AA"
console.log(gcdOfStringsOptimized("AA", "A")); // "A"
General Problem-Solving Strategies:
- Understand the Problem: Carefully read the problem statement to understand the requirements and constraints.
- Break Down the Problem: Identify key operations such as checking concatenations and computing GCD of lengths.
-
Use Helper Functions: Implement helper functions like
gcd
to simplify the main solution. - Consider Edge Cases: Think about different scenarios that might affect the solution, such as when the strings are not divisible or one string is empty.
- Start Simple and Optimize: Begin with a basic solution that works, even if it’s not the most efficient. This helps to ensure you understand the problem. Then, look for ways to optimize it.
- Optimize for Readability: Ensure the solution is clean and easy to understand while maintaining efficiency.
- Test Thoroughly: Test your solution with various cases, including edge cases. Ensure that the solution handles all possible inputs correctly.
Identifying Similar Problems:
String Repetition and Pattern Matching:
- Problems where you need to find repeating patterns within strings.
- Example: Finding the smallest repeating unit in a string.
Greatest Common Divisor (GCD):
- Problems involving GCD computations in different contexts.
- Example: Finding the GCD of two numbers or the GCD of lengths of multiple strings.
String Concatenation and Validation:
- Problems where you need to validate if one string can be formed by concatenating another string multiple times.
- Example: Checking if a string is a repeated substring of another string.
Subsequence and Substring Problems:
- Problems involving finding common subsequences or substrings between multiple strings.
- Example: Longest common subsequence or longest common substring problems.
Conclusion:
- The problem of finding the greatest common divisor of two strings can be efficiently solved by leveraging string concatenation and GCD computations.
- Understanding the problem and breaking it down into manageable parts is crucial.
- Testing with various edge cases ensures robustness.
- Recognizing patterns in problems can help apply similar solutions to other challenges.
By practicing such problems and strategies, you can improve your problem-solving skills and be better prepared for various coding challenges.
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