🧙‍♂️Beginner-Friendly Guide "Find the K-th Character in String Game I" – LeetCode 3304 (C++ | Python | JavaScript)

#programming #cpp #javascript #python

Hey adventurers!

In this playful string-based puzzle, we join Alice and Bob in an ever-growing word game. From just a single letter, a mysterious pattern evolves — and you’re tasked with figuring out what the k-th character becomes after repeated transformations. 🌀

Let’s demystify it together.

🧠 Problem Summary

You're given:

A game that starts with the string word = "a"
A number k, indicating the position of the character you want to retrieve

Each round:

Every character in the current word is changed to its next character in the alphabet, and the result is appended to the word.

Examples:

"a" becomes "ab"
"ab" becomes "abbc"
"abbc" becomes "abbcbccd"

Your goal:

Return the character at position k (1-based index) after enough rounds.

💡 Intuition

If you trace the pattern carefully, you'll realize:

The transformation is deterministic.
Each new character in the word is one step ahead of its source.
The position of each new character follows a binary-like structure — much like the count of 1s in the binary representation of k-1.

Thus, the answer is simply:

'a' + popcount(k - 1)

Where popcount(x) counts the number of 1s in the binary representation of x.

🛠️ C++ Code

class Solution {
 public:
  char kthCharacter(unsigned k) {
    return 'a' + popcount(k - 1);
  }
};

🐍 Python Code

class Solution:
    def kthCharacter(self, k: int) -> str:
        return chr(ord('a') + bin(k - 1).count('1'))

💻 JavaScript Code

var kthCharacter = function(k) {
    const popcount = n => n.toString(2).split('1').length - 1;
    return String.fromCharCode('a'.charCodeAt(0) + popcount(k - 1));
};

📝 Key Insights

This elegant transformation can be reduced to analyzing binary patterns.
Think of each bit in k-1 as a transformation step.
No need to simulate string growth — bitwise logic wins here. 🧠

✅ Final Thoughts

From a simple "a" to a cascade of characters, this problem rewards observation and pattern recognition. It's a neat reminder that clever math can beat brute force.

If you enjoyed this one, share it with a fellow coder! Happy decoding! 💻✨