Given two strings **s1** and **s2**, return their longest common prefix

**Example 1:**

Input:s1 = "hello", s2 = "help"Output:"hel"

**Example 2:**

Input:s1 = "Andy", s2 = "Mircea"Output:""

**Example 3:**

Input:s1 = "AlgoCademy", s2 = "Algo"Output:"Algo"

The core challenge of this problem is to find the longest common prefix between two given strings. This is a common problem in string manipulation and has applications in text processing, DNA sequence analysis, and more. A potential pitfall is assuming that the common prefix will always be non-empty, which is not the case as shown in Example 2.

To solve this problem, we can start by comparing characters of both strings one by one from the beginning until we find a mismatch or reach the end of one of the strings. This approach ensures that we find the longest common prefix efficiently.

A naive solution would involve nested loops to compare each character of the first string with each character of the second string. However, this is not optimal and has a time complexity of O(n*m), where n and m are the lengths of the two strings.

An optimized solution involves a single loop to compare characters of both strings until a mismatch is found. This approach has a time complexity of O(n), where n is the length of the shorter string.

Here is a step-by-step breakdown of the optimized algorithm:

- Initialize an index variable to 0.
- Loop through both strings until the characters at the current index are the same and the index is less than the length of both strings.
- Increment the index variable.
- Return the substring of the first string from the beginning to the current index.

```
public class LongestCommonPrefix {
public static String longestCommonPrefix(String s1, String s2) {
int minLength = Math.min(s1.length(), s2.length());
int i = 0;
// Loop until characters match and within the bounds of the shorter string
while (i < minLength && s1.charAt(i) == s2.charAt(i)) {
i++;
}
// Return the common prefix
return s1.substring(0, i);
}
public static void main(String[] args) {
// Test cases
System.out.println(longestCommonPrefix("hello", "help")); // Output: "hel"
System.out.println(longestCommonPrefix("Andy", "Mircea")); // Output: ""
System.out.println(longestCommonPrefix("AlgoCademy", "Algo")); // Output: "Algo"
}
}
```

The time complexity of the optimized solution is O(n), where n is the length of the shorter string. The space complexity is O(1) as we are using a constant amount of extra space.

Potential edge cases include:

- One or both strings are empty: The output should be an empty string.
- No common prefix: The output should be an empty string.
- One string is a prefix of the other: The output should be the shorter string.

Examples:

Input:s1 = "", s2 = "hello"Output:""Input:s1 = "hello", s2 = ""Output:""Input:s1 = "prefix", s2 = "pre"Output:"pre"

To test the solution comprehensively, consider the following test cases:

- Both strings are empty.
- One string is empty.
- No common prefix.
- One string is a prefix of the other.
- Both strings are identical.

Using a testing framework like JUnit can help automate and validate these test cases.

When approaching string problems, always consider edge cases and constraints. Breaking down the problem into smaller steps and thinking about the simplest solution first can help in deriving an optimized solution. Practice similar problems to improve your problem-solving skills.

Understanding and solving the longest common prefix problem is crucial for various applications in text processing and other fields. By practicing and exploring different approaches, you can enhance your problem-solving skills and tackle more complex problems effectively.