Typescript Coding Chronicles: Greatest Common Divisor of Strings
Tags: typescript, webdev, programming, javascript
Published: 2024-07-10 18:22:46

## 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` and `str2` 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:
```typescript
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 of `str2`. 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:
```typescript
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 of `str2`. 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:
1. `str1` is a multiple of `str2`.
2. `str2` is a multiple of `str1`.
3. `str1` and `str2` have no common divisor.
4. `str1` or `str2` is empty.

### Test Cases:
```typescript
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:
1. **Understand the Problem:** Carefully read the problem statement to understand the requirements and constraints.
2. **Break Down the Problem:** Identify key operations such as checking concatenations and computing GCD of lengths.
3. **Use Helper Functions:** Implement helper functions like `gcd` to simplify the main solution.
4. **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.
5. **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.
6. **Optimize for Readability:** Ensure the solution is clean and easy to understand while maintaining efficiency.
7. **Test Thoroughly:** Test your solution with various cases, including edge cases. Ensure that the solution handles all possible inputs correctly.

## Identifying Similar Problems:
1. **String Repetition and Pattern Matching:**
   - Problems where you need to find repeating patterns within strings.
   - Example: Finding the smallest repeating unit in a string.

2. **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.

3. **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.

4. **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.