Problem Description

Alice and Bob are playing a game where Alice starts with a string word = "a". Given a positive integer k, Bob asks Alice to repeatedly perform a specific operation on the string.

The operation works as follows:

• Take the current word string
• Change each character to its next character in the English alphabet (a→b, b→c, ..., z→a)
• Append this transformed string to the original word

For example:

• Starting with "c", the operation produces "cd" (c becomes d, append to get "cd")
• Starting with "zb", the operation produces "zbac" (z→a, b→c, append to get "zbac")

The game continues with Alice performing this operation repeatedly until the string has at least k characters. Your task is to find and return the kth character (1-indexed) in the final string.

The solution simulates this process by maintaining an array where each element represents the offset from 'a' for each character. Starting with [0] (representing "a"), it repeatedly extends the array by adding the incremented values (modulo 26 for wraparound) until the array has at least k elements. The kth character is then converted back to its corresponding letter.

Intuition

The key observation is that we need to simulate the string growth process until we have at least k characters. Since we're only interested in the kth character, we don't need to store the actual string - we can work with numerical representations.

Think about what happens in each operation:

• We take the current string
• Transform each character to its next letter
• Append the transformed version to the original

This means the string doubles in length with each operation. Starting with "a" (length 1), after one operation we have "ab" (length 2), then "abbc" (length 4), then "abbcbccd" (length 8), and so on.

Instead of working with actual characters, we can represent each character as its offset from 'a'. So 'a' = 0, 'b' = 1, 'c' = 2, and so on. This makes the transformation operation simple arithmetic: just add 1 to each value (using modulo 26 to handle the wraparound from 'z' to 'a').

The pattern becomes clear:

• Start with [0] representing "a"
• First operation: append [1] to get [0, 1] representing "ab"
• Second operation: append [1, 2] to get [0, 1, 1, 2] representing "abbc"
• Continue until we have at least k elements

Since we only need the kth character, we can stop as soon as our array reaches length k, then convert the value at index k-1 back to its corresponding character by adding it to the ASCII value of 'a'.

Learn more about Recursion and Math patterns.

Solution Approach

The simulation approach uses an array to track the character offsets from 'a' and builds the string iteratively until we have enough characters.

Implementation Steps:

1. Initialize the word array: Start with word = [0], which represents the initial string "a" where 0 is the offset from 'a'.

2. Grow the array through operations: While the length of word is less than k, perform the operation:

   • Take the current word array
   • Create a new array by incrementing each value: [(x + 1) % 26 for x in word]
   • The modulo 26 handles the wraparound (z → a)
   • Extend the original array with these new values using word.extend()

3. Extract the kth character: Once we have at least k elements:

   • Access the element at index k - 1 (converting from 1-indexed to 0-indexed)
   • Convert the numerical offset back to a character: chr(ord("a") + word[k - 1])

Example walkthrough with k = 5:

• Initial: word = [0] (represents "a")
• After operation 1: word = [0, 1] (represents "ab")
• After operation 2: word = [0, 1, 1, 2] (represents "abbc")
• After operation 3: word = [0, 1, 1, 2, 1, 2, 2, 3] (represents "abbcbccd")
• Now len(word) >= 5, so we return chr(ord("a") + word[4]) = chr(97 + 1) = 'b'

The time complexity is O(k) since we need to generate at least k characters, and the space complexity is also O(k) for storing the array.

Example Walkthrough

Let's trace through the solution with k = 10 to find the 10th character.

Initial State:

• word = [0] representing "a"
• Length = 1, need to reach at least 10

Operation 1:

• Current: [0]
• Transform each element: 0 + 1 = 1
• Append [1] to get [0, 1]
• This represents "ab"
• Length = 2

Operation 2:

• Current: [0, 1]
• Transform: [0+1, 1+1] = [1, 2]
• Append to get [0, 1, 1, 2]
• This represents "abbc"
• Length = 4

Operation 3:

• Current: [0, 1, 1, 2]
• Transform: [1, 2, 2, 3]
• Append to get [0, 1, 1, 2, 1, 2, 2, 3]
• This represents "abbcbccd"
• Length = 8

Operation 4:

• Current: [0, 1, 1, 2, 1, 2, 2, 3]
• Transform: [1, 2, 2, 3, 2, 3, 3, 4]
• Append to get [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4]
• This represents "abbcbccdbccdbcde"
• Length = 16

Finding the 10th character:

• We now have 16 elements (≥ 10)
• Access index k-1 = 9: word[9] = 2
• Convert to character: chr(ord('a') + 2) = chr(97 + 2) = 'c'

The 10th character is 'c'.

To verify: the string "abbcbccdbccdbcde" has 'c' at position 10 (1-indexed).

Solution Implementation

Time and Space Complexity

Time Complexity: O(k)

The algorithm starts with a list containing one element [0]. In each iteration of the while loop, it doubles the size of the list by extending it with modified values. The loop continues until len(word) >= k.

• Initial size: 1
• After iteration 1: 2
• After iteration 2: 4
• After iteration 3: 8
• After iteration i: 2^i

The loop runs approximately log₂(k) iterations to reach size k. However, in each iteration, the extend operation processes all current elements to create the new elements. The total work done across all iterations is:

• Iteration 1: process 1 element
• Iteration 2: process 2 elements
• Iteration 3: process 4 elements
• ...

Total work = 1 + 2 + 4 + ... + 2^(log₂(k)-1) = 2^(log₂(k)) - 1 ≈ k - 1

Therefore, the time complexity is O(k).

Space Complexity: O(k)

The algorithm creates a list word that grows until its size is at least k. In the worst case, the list size will be the smallest power of 2 that is greater than or equal to k, which is at most 2k. Therefore, the space complexity is O(k).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Inefficient String Concatenation

A common mistake is using string concatenation directly instead of working with numeric representations:

Problematic approach:

def kthCharacter(self, k: int) -> str:
    word = "a"
    while len(word) < k:
        new_part = ""
        for char in word:
            if char == 'z':
                new_part += 'a'
            else:
                new_part += chr(ord(char) + 1)
        word += new_part
    return word[k-1]

Why it's inefficient: String concatenation in Python creates new string objects each time, leading to O(k²) time complexity due to repeated string copying.

Solution: Use the numeric array approach as shown in the original solution, or use a list of characters and join at the end.

2. Forgetting the Wraparound from 'z' to 'a'

Without proper modulo operation, the code might fail when encountering 'z':

Incorrect:

new_portion = [char_value + 1 for char_value in word]  # Missing % 26

This would produce values beyond 25, causing incorrect characters or runtime errors.

Solution: Always use (char_value + 1) % 26 to ensure values stay within the valid range [0, 25].

3. Off-by-One Error with Indexing

Confusing 0-indexed and 1-indexed positions is a frequent mistake:

Incorrect:

return chr(ord('a') + word[k])  # Should be word[k-1]

Since the problem asks for the k-th character (1-indexed), we need to access word[k-1] in our 0-indexed array.

4. Unnecessary Full String Generation

Some might generate the entire final string even when only one character is needed:

Inefficient for large k:

def kthCharacter(self, k: int) -> str:
    word = "a"
    while len(word) < k:
        # ... operation ...
    # Converting entire array to string unnecessarily
    result_string = ''.join([chr(ord('a') + x) for x in word])
    return result_string[k-1]

Solution: Only convert the specific character needed: chr(ord('a') + word[k-1])

5. Not Handling Edge Cases

While the problem guarantees k is positive, defensive programming would check:

Better practice:

if k == 1:
    return 'a'  # Quick return for the base case

This optimization avoids unnecessary iterations when k=1.