Longest Common Prefix

1. Defining the Problem

First, let's ensure we have a clear understanding of the task. The goal is to identify the longest prefix shared by all strings in an array. If no such prefix exists (e.g., in ["dog", "racecar", "car"]), the result should be an empty string "".

The "takeuforward" article, which is part of the TUF DSA sheet you're following, provides a concise definition and examples.

2. Approach 1: Brute-Force (Vertical Scanning)

The most direct way to solve this is to compare the strings character by character, vertically. Imagine the strings are listed one below the other. We check the first character of all strings. If they all match, we append it to our result and move to the second character. We continue this process until we find a mismatch or one of the strings ends.

The logic is as follows:

1. Iterate through the characters of the first string, using an index i.
2. For each character c at strs[0].charAt(i), start a second loop to iterate through all the other strings in the array (from strs[1] onwards).
3. In the inner loop, check two conditions:
   • Has the current string (strs[j]) become too short (i.e., i is out of bounds)?
   • Does the character strs[j].charAt(i) not match c?
4. If either of these is true, we have found the end of the common prefix. The answer is the substring of the first string from the beginning up to the current index i.
5. If the loops complete without any mismatches, it means the entire first string is the common prefix.

This method is often called "Vertical Scanning" because we are effectively scanning down the "columns" of characters at each index. The following video provides an excellent, detailed walkthrough of this exact approach, including a line-by-line debugging session.

The time complexity is roughly $O(N\times M)$, where $N$ is the number of strings and $M$ is the length of the longest common prefix. This is because, in the best case, we check $M$ characters for all $N$ strings.

3. Approach 2: A Better Method (Horizontal Scanning)

An alternative approach is to work "horizontally." Instead of comparing character by character across strings, we can take the entire first string as our initial candidate for the prefix and then shorten it as needed.

The logic works like this:

1. Initialize the prefix variable with the first string in the array (strs[0]).
2. Iterate through the rest of the strings in the array, from strs[1] onwards.
3. For each string, check if it starts with the current prefix. In Java, string.indexOf(prefix) != 0 is a common way to do this.
4. If it doesn't, shorten the prefix by removing the last character. Repeat this until the string starts with the prefix or the prefix becomes empty.
5. If the prefix becomes empty at any point, we can stop early and return "".
6. The prefix that survives this process for all strings is the final answer.

This "Horizontal Scanning" method is intuitive and easy to implement. The video below explains this concept clearly.

The time complexity of this approach is $O(S)$, where $S$ is the sum of all characters in all strings. In the worst case, where all strings are nearly identical, this can be quite inefficient as the indexOf operation repeatedly scans the strings. However, it's a valid and common approach you should be aware of.

4. Approach 3: An Optimal and Elegant Solution (Sorting)

Now, let's explore a particularly clever solution that aligns with the Striver DSA sheet. This approach leverages a key insight about lexicographical ordering.

If you sort an array of strings alphabetically, the longest common prefix for the entire array must be contained within the common prefix of the very first and very last strings in the sorted array. Why? Because the first and last elements are the ones that are most different from each other. Any prefix they share must also be shared by all the strings that lie between them.

This simplifies the problem dramatically:

1. Sort the input array of strings.
2. Compare just two strings: the first one and the last one.
3. Their common prefix is the answer for the entire array.

This is a powerful technique that reduces a problem of comparing N strings to a problem of comparing just two strings, with the main cost being the initial sort. Let's study the implementation from the takeuforward resource.

The time complexity is dominated by the sorting step. The resource gives it as $O(N\cdot \log N+M)$. This assumes the cost of comparing two strings is constant. A more precise analysis would be $O(L\cdot N\cdot \log N)$, where $L$ is the average string length, because each comparison during the sort takes time proportional to the string length. Nonetheless, the sorting-based approach is very efficient and demonstrates a creative problem-solving mindset highly valued in interviews. The space complexity is typically $O(\log N)$ or $O(N)$ depending on the implementation of the sorting algorithm, plus $O(M)$ for storing the resulting prefix.

Conclusion

In this lesson, we explored three different ways to find the longest common prefix. Each approach reveals a different way of thinking about the problem, a skill that is crucial for technical interviews.

Key Takeaways:

• Vertical Scanning: A direct, brute-force method that compares characters column by column. It's simple to understand and implement.
• Horizontal Scanning: An iterative approach that assumes a prefix and shortens it. It provides an alternative perspective on the problem.
• Sorting Approach: The most elegant and often optimal solution. It relies on the clever insight that sorting the array reduces the problem to a comparison between just the first and last elements.

You've now added another important string problem to your toolkit. In our next lesson, we will pivot to a powerful new pattern for solving string and array problems: the sliding window technique. We'll start by applying it to problems with a fixed-size window, such as finding the maximum sum of a subarray of a specific size.