LeetCode #3304 — EASY

Find the K-th Character in String Game I

Build confidence with an intuition-first walkthrough focused on math fundamentals.

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The Problem

Problem Statement

Alice and Bob are playing a game. Initially, Alice has a string word = "a".

You are given a positive integer k.

Now Bob will ask Alice to perform the following operation forever:

Generate a new string by changing each character in word to its next character in the English alphabet, and append it to the original word.

For example, performing the operation on "c" generates "cd" and performing the operation on "zb" generates "zbac".

Return the value of the k^th character in word, after enough operations have been done for word to have at least k characters.

Example 1:

Input: k = 5

Output: "b"

Explanation:

Initially, word = "a". We need to do the operation three times:

Generated string is "b", word becomes "ab".
Generated string is "bc", word becomes "abbc".
Generated string is "bccd", word becomes "abbcbccd".

Example 2:

Input: k = 10

Output: "c"

Constraints:

1 <= k <= 500

Step 01

Brute Force Baseline

Problem summary: Alice and Bob are playing a game. Initially, Alice has a string word = "a". You are given a positive integer k. Now Bob will ask Alice to perform the following operation forever: Generate a new string by changing each character in word to its next character in the English alphabet, and append it to the original word. For example, performing the operation on "c" generates "cd" and performing the operation on "zb" generates "zbac". Return the value of the kth character in word, after enough operations have been done for word to have at least k characters.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Bit Manipulation

Example 1

Example 2

Core Insight

What unlocks the optimal approach

The constraints are small. Construct the string by simulating the operations.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3304: Find the K-th Character in String Game I
class Solution {
    public char kthCharacter(int k) {
        List<Integer> word = new ArrayList<>();
        word.add(0);
        while (word.size() < k) {
            for (int i = 0, m = word.size(); i < m; ++i) {
                word.add((word.get(i) + 1) % 26);
            }
        }
        return (char) ('a' + word.get(k - 1));
    }
}

// Accepted solution for LeetCode #3304: Find the K-th Character in String Game I
func kthCharacter(k int) byte {
	word := []int{0}
	for len(word) < k {
		m := len(word)
		for i := 0; i < m; i++ {
			word = append(word, (word[i]+1)%26)
		}
	}
	return 'a' + byte(word[k-1])
}

# Accepted solution for LeetCode #3304: Find the K-th Character in String Game I
class Solution:
    def kthCharacter(self, k: int) -> str:
        word = [0]
        while len(word) < k:
            word.extend([(x + 1) % 26 for x in word])
        return chr(ord("a") + word[k - 1])

// Accepted solution for LeetCode #3304: Find the K-th Character in String Game I
impl Solution {
    pub fn kth_character(k: i32) -> char {
        let mut word = vec![0];
        while word.len() < k as usize {
            let m = word.len();
            for i in 0..m {
                word.push((word[i] + 1) % 26);
            }
        }
        (b'a' + word[(k - 1) as usize] as u8) as char
    }
}

// Accepted solution for LeetCode #3304: Find the K-th Character in String Game I
function kthCharacter(k: number): string {
    const word: number[] = [0];
    while (word.length < k) {
        word.push(...word.map(x => (x + 1) % 26));
    }
    return String.fromCharCode(97 + word[k - 1]);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(k)

Space

O(k)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.