LeetCode 3307: Find the K-th Character in String Game II Solution

Master LeetCode problem 3307 (Find the K-th Character in String Game II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3307. Find the K-th Character in String Game II

Hard

Math Bit Manipulation Recursion

Problem Explanation

Explanation

To solve this problem, we need to simulate the operations that Alice performs on the initial string. We can keep track of the length of the string after each operation and find the k-th character based on the number of operations performed. We can use modulo arithmetic to determine the position of the k-th character in the final string.

Algorithmic Idea

Initialize the string word with "a" and set the current length of the word to 1.
Iterate over the operations array and perform the specified operation.
For operation 0, double the length of the word.
For operation 1, increment each character in the word by one and append the resulting string to the original word.
Calculate the effective length of the word after all operations.
Find the k-th character in the final word using modulo arithmetic.

Time Complexity

The time complexity of this algorithm is O(n), where n is the length of the operations array.

Space Complexity

The space complexity of this algorithm is O(n), where n is the length of the operations array.

Solution Code

class Solution {
    public char findKthCharacter(int k, int[] operations) {
        String word = "a";
        int length = 1;
        
        for (int operation : operations) {
            if (operation == 0) {
                length *= 2;
            } else {
                StringBuilder sb = new StringBuilder();
                for (char ch : word.toCharArray()) {
                    sb.append((char) ((ch - 'a' + 1) % 26 + 'a'));
                }
                word += sb.toString();
            }
        }
        
        int effectiveLength = word.length();
        
        k--; // Convert k to 0-based index
        k %= effectiveLength;
        
        return word.charAt(k);
    }
}

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