The k-th Lexicographical String of All Happy Strings of Length n

The k-th Lexicographical String of All Happy Strings of Length n

What is this problem about?

Combinatorial problems often involve generating strings that follow specific constraints. A "happy string" is defined as a string consisting only of the characters 'a', 'b', and 'c', where no two adjacent characters are identical. For a given length 'n', there are a finite number of such happy strings. When these strings are arranged in alphabetical (lexicographical) order, the task is to find the one that occupies the 'k-th' position. If 'k' is larger than the total number of possible happy strings, the solution should return an empty string. This problem requires a blend of counting logic and string construction.

Why is this asked in interviews?

This the k-th lexicographical string of all happy strings of length n interview question is frequently used by top-tier tech firms like Microsoft, Meta, Amazon, and Google. It assesses a candidate's ability to navigate state spaces using backtracking and their grasp of combinatorial mathematics. Specifically, it tests whether you can avoid generating every possible solution (which could be computationally expensive) by mathematically determining which branch of the decision tree the 'k-th' string resides in. This "counting to find the k-th" technique is a high-level algorithmic skill highly valued in engineering.

Algorithmic pattern used

This problem primarily utilizes the Backtracking, String interview pattern. You can think of the generation process as building a tree where each node is a character. The root has 3 choices ('a', 'b', 'c'). Each subsequent node has 2 choices (the two characters not used by the parent). For a string of length 'n', the total number of happy strings is 3 * 2^(n-1). By knowing the number of strings that can be formed starting with a certain prefix, you can determine if the 'k-th' string is in the current subtree. If 'k' is less than or equal to the count of strings in a subtree, you move into it. Otherwise, you subtract that count from 'k' and skip to the next available character at the same level.

Example explanation

Suppose n = 2 and we want the k = 4th happy string. The possible strings are: "ab", "ac", "ba", "bc", "ca", "cb".

1. Start with the first character. Options are 'a', 'b', 'c'.
2. If we start with 'a', how many happy strings of length 2 exist? Since the second char can be 'b' or 'c', there are 2 strings.
3. Our target is k = 4. Since 4 > 2, we know the string does not start with 'a'. Update k = 4 - 2 = 2.
4. Next option is 'b'. How many strings start with 'b'? There are 2 ("ba", "bc").
5. Our updated k = 2. Since 2 <= 2, the string must start with 'b'.
6. Now for the second character. Options are 'a' and 'c' (not 'b').
7. How many strings start with "ba"? Just 1.
8. Since k = 2 and 2 > 1, the second char is not 'a'. Update k = 2 - 1 = 1.
9. Next option is 'c'. The string is "bc". Wait, in my manual list "bc" was the 4th string. Let's re-verify: "ab"(1), "ac"(2), "ba"(3), "bc"(4). Correct.

Common mistakes candidates make

In "The k-th lexicographical string of all happy strings of length n coding problem," candidates often try to generate all strings and store them in a list before picking the k-th. While this works for small 'n', it is highly inefficient for larger values and shows a lack of optimization skill. Another error is incorrectly calculating the number of possible strings in each branch, leading to "off-by-one" errors in the selection process. Finally, many forget to handle the case where 'k' exceeds the total number of possible happy strings.

Interview preparation tip

Mastering backtracking requires practice with recursion and tree-based visualization. When dealing with lexicographical problems, always ask yourself if you can "skip" branches of the search tree by counting. This technique is applicable to many "k-th permutation" or "k-th combination" problems. Understanding powers of two and basic combinatorics will help you solve these types of string problems much faster.