String Compression III

String Compression III

What is this problem about?

String Compression III is a medium-level problem that asks you to compress a string using a specific set of rules. Unlike basic run-length encoding, this version often introduces a constraint on the maximum length of a single compressed segment. For instance, you might be asked to compress a string such that no single character count exceeds 9. If a character repeats more than 9 times, it must be split into multiple segments. For example, "aaaaaaaaaaaa" (12 'a's) would be compressed as "9a3a". The objective is to iterate through the string and build the compressed version by counting consecutive identical characters while respecting these limits.

Why is this asked in interviews?

This question is a favorite at companies like Apple, Microsoft, and Google because it tests fundamental string traversal and logic implementation. While it's easier than its dynamic programming counterpart (String Compression II), it requires careful handling of edge cases and boundary conditions. Interviewers use it to see if a candidate can write clean, efficient code that handles "chunking" logic correctly. It's a great way to assess if someone can translate a set of rules into a working algorithm without overcomplicating the solution.

Algorithmic pattern used

The core pattern used here is the Two Pointers or the simple Linear Scan. You use one pointer to track the current character being processed and another (or a nested loop) to count how many times that character repeats consecutively. Once the count reaches the defined limit (like 9) or the character changes, you append the count and the character to the result string and move on. This ensures that the problem is solved in O(n) time complexity, making it highly efficient.

Example explanation (use your own example)

Let's take the string "vvvvvvvvvvvvvzz".

1. We start with the first character 'v'.
2. We count consecutive 'v's. There are 13.
3. Since our limit is 9, we take 9 'v's and append "9v" to our result.
4. We have 4 'v's left. We append "4v" to our result.
5. Next, we see 'z'. We count consecutive 'z's. There are 2.
6. We append "2z" to our result. The final compressed string is "9v4v2z".

Common mistakes candidates make

A frequent error is forgetting to handle the end of the string, which can lead to the last group of characters not being added to the result. Another mistake is not resetting the counter correctly after reaching the maximum segment limit (e.g., 9). Some candidates might also try to use complex data structures like HashMaps when a simple character count is sufficient, leading to unnecessary memory overhead. Finally, off-by-one errors in the loop condition are common when navigating the indices.

Interview preparation tip

When preparing for the String Compression III coding problem, practice writing the logic using a single while loop or nested while loops to see which approach feels more intuitive for you. Always clarify the constraints with your interviewer—specifically, the maximum segment length. Making sure your code handles strings with only one character or very long strings with no repeating characters is a great way to show attention to detail.