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Find the K-th Character in String Game I

Easy

Asked by:

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Topics:

Strings

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

Try coding it in LeetCode

Find the K-th Character in String Game I

Easy

Asked by:

93 views

Topics:

Strings

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range for `n` and `k`? Can they be zero or negative?
Can the input string `s` be empty or contain characters other than '0' and '1'?
If `k` is larger than the length of the fully expanded string, what should I return? Should I return an error or a specific value?
Could you provide an example where the fully expanded string is very long, so I can better understand the intended behavior and edge cases?
Is there a specific character encoding I should be aware of (e.g., ASCII, UTF-8) that might affect character comparisons or manipulation?

Brute Force Solution

Approach

The brute force method for finding a character involves creating the string in full, according to the rules, until we reach the character we want. We keep expanding the string step by step. Then, we directly access that specific character.

Here's how the algorithm would work step-by-step:

Start with the initial string.
If the desired character position is within the current string, we're done.
Otherwise, create the next version of the string based on the current one, according to the defined transformation rules.
Repeat the creation process, checking after each step if the desired character position is within the newly created string.
Continue this process until the length of the generated string is at least as long as the position of the character you're looking for.
Once the string is long enough, pick out the character at the requested position.

Code Implementation

def find_kth_character_brute_force(initial_string, k_position):
    current_string = initial_string

    while True:
        # Check if the k_position is within the current string length
        if k_position <= len(current_string):
            break

        # Create the next version of the string based on current string
        next_string = current_string + current_string
        current_string = next_string

        # Keep looping until the current string is at least as long as k_position

    # Once the string is long enough, return the character at k_position - 1 (0-indexed)
    return current_string[k_position - 1]

Big(O) Analysis

Time Complexity

O(2^K) – The brute force method constructs the string step by step until the K-th character is reached. In each step, the string doubles in size. So, to reach the K-th character, we need to expand the string until its length is at least K. Since the length doubles in each iteration, the length of the string grows as 2^0, 2^1, 2^2, and so on until it reaches at least K. The total time complexity is proportional to the sum of the lengths of these strings in each expansion, which is O(1 + 2 + 4 + ... + K). This sum is approximately equal to the final length K, but the creation of the final length dominates. More accurately, it takes O(2^K) time to build a string of length K by doubling. The final character lookup is O(1) and thus is insignificant.

Space Complexity

O(N) – The brute force method described repeatedly creates the next version of the string. The space complexity is driven by storing this growing string. In the worst case, the string will have to grow to at least the size of the desired character position K. Therefore, the auxiliary space required is proportional to K, which can be considered N for the purposes of space complexity analysis, leading to O(N) space.

Optimal Solution

Approach

We don't need to actually construct the entire transformed string. The key idea is to track how the string grows with each transformation and efficiently jump to the relevant transformation step using math. This lets us figure out the character at the requested position without simulating all the steps.

Here's how the algorithm would work step-by-step:

Figure out the size of the string after each transformation, and see if the position of the character we want is even within this string.
If the desired position is outside of the current string size, this means the character at the desired position stays the same.
If the desired position falls inside of the string, determine the transformation step that created that character.
The character we are looking for in the transformed string is derived from a character in the earlier string. Essentially, determine which character in the original string was transformed to become the character in the transformed string.
Continue tracing the character backwards through transformations until we arrive at the original string. Return the character in the original string at the initial position.

Code Implementation

def find_kth_character(original_string, position, transformation_count):
    string_length = len(original_string)

    for _ in range(transformation_count):
        transformed_string_length = 2 * string_length - 1

        if position > transformed_string_length:
            continue

        # If position is beyond string length, the char stays the same
        if position > string_length:
            position = transformed_string_length - position + 1
        else:
            # Current pos depends on char from earlier string
            return original_string[position - 1]

        string_length = transformed_string_length

    return original_string[position - 1]

Big(O) Analysis

Time Complexity

O(log n) – The algorithm does not construct the transformed string. The dominant cost comes from the while loop that traces the character position backward through the transformations. In each iteration, the algorithm checks if the position falls within the current string length and if so, it effectively halves the string length by determining which position in the prior string contributed to it. This halving process results in a logarithmic number of iterations with respect to the initial position, n. The operations inside the loop are constant time operations. Therefore, the time complexity is O(log n).

Space Complexity

O(1) – The algorithm iteratively traces back through the transformations to find the original character. It doesn't use any auxiliary data structures like arrays, lists, or hash maps to store intermediate results or visited states. It only uses a few variables to track the position as it traces back. Therefore, the space used is constant and does not depend on the input string length or the transformation steps.

Edge Cases

Case	How to Handle
Empty string s or invalid n (n <= 0)	Return empty string or throw exception, respectively, as no Kth character is possible.
K is out of bounds (K < 0 or K >= length of generated string)	Return empty string or throw exception since the Kth character does not exist.
n is very large, leading to potentially exponential growth and stack overflow.	Iterative approach is preferred over recursive to avoid stack overflow issues with large n.
s contains non-alphanumeric characters	Specify behavior (e.g., strip non-alphanumeric characters or throw an exception if not allowed).
s contains upper and lower case characters	Define whether the comparison is case-sensitive or case-insensitive during string replacement.
String s is very long, impacting memory usage when generating strings.	Optimize string concatenation to minimize memory allocation overhead and use StringBuilder in Java.
The generated string before reaching n is shorter than K.	Return null or throw an exception because a character at index K doesn't exist in the generated string.
Integer overflow when calculating length of repeated string.	Use appropriate data types (e.g., long) to prevent integer overflow or implement length check prior to string building.

Empty string s or invalid n (n <= 0)

How to Handle:

Return empty string or throw exception, respectively, as no Kth character is possible.

K is out of bounds (K < 0 or K >= length of generated string)

How to Handle:

Return empty string or throw exception since the Kth character does not exist.

n is very large, leading to potentially exponential growth and stack overflow.

How to Handle:

Iterative approach is preferred over recursive to avoid stack overflow issues with large n.

s contains non-alphanumeric characters

How to Handle:

Specify behavior (e.g., strip non-alphanumeric characters or throw an exception if not allowed).

s contains upper and lower case characters

How to Handle:

Define whether the comparison is case-sensitive or case-insensitive during string replacement.

String s is very long, impacting memory usage when generating strings.

How to Handle:

Optimize string concatenation to minimize memory allocation overhead and use StringBuilder in Java.

The generated string before reaching n is shorter than K.

How to Handle:

Return null or throw an exception because a character at index K doesn't exist in the generated string.

Integer overflow when calculating length of repeated string.

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflow or implement length check prior to string building.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range for `n` and `k`? Can they be zero or negative?
Can the input string `s` be empty or contain characters other than '0' and '1'?
If `k` is larger than the length of the fully expanded string, what should I return? Should I return an error or a specific value?
Could you provide an example where the fully expanded string is very long, so I can better understand the intended behavior and edge cases?
Is there a specific character encoding I should be aware of (e.g., ASCII, UTF-8) that might affect character comparisons or manipulation?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the initial string.
If the desired character position is within the current string, we're done.
Otherwise, create the next version of the string based on the current one, according to the defined transformation rules.
Repeat the creation process, checking after each step if the desired character position is within the newly created string.
Continue this process until the length of the generated string is at least as long as the position of the character you're looking for.
Once the string is long enough, pick out the character at the requested position.

Code Implementation

def find_kth_character_brute_force(initial_string, k_position):
    current_string = initial_string

    while True:
        # Check if the k_position is within the current string length
        if k_position <= len(current_string):
            break

        # Create the next version of the string based on current string
        next_string = current_string + current_string
        current_string = next_string

        # Keep looping until the current string is at least as long as k_position

    # Once the string is long enough, return the character at k_position - 1 (0-indexed)
    return current_string[k_position - 1]

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Figure out the size of the string after each transformation, and see if the position of the character we want is even within this string.
If the desired position is outside of the current string size, this means the character at the desired position stays the same.
If the desired position falls inside of the string, determine the transformation step that created that character.
The character we are looking for in the transformed string is derived from a character in the earlier string. Essentially, determine which character in the original string was transformed to become the character in the transformed string.
Continue tracing the character backwards through transformations until we arrive at the original string. Return the character in the original string at the initial position.

Code Implementation

def find_kth_character(original_string, position, transformation_count):
    string_length = len(original_string)

    for _ in range(transformation_count):
        transformed_string_length = 2 * string_length - 1

        if position > transformed_string_length:
            continue

        # If position is beyond string length, the char stays the same
        if position > string_length:
            position = transformed_string_length - position + 1
        else:
            # Current pos depends on char from earlier string
            return original_string[position - 1]

        string_length = transformed_string_length

    return original_string[position - 1]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty string s or invalid n (n <= 0)	Return empty string or throw exception, respectively, as no Kth character is possible.
K is out of bounds (K < 0 or K >= length of generated string)	Return empty string or throw exception since the Kth character does not exist.
n is very large, leading to potentially exponential growth and stack overflow.	Iterative approach is preferred over recursive to avoid stack overflow issues with large n.
s contains non-alphanumeric characters	Specify behavior (e.g., strip non-alphanumeric characters or throw an exception if not allowed).
s contains upper and lower case characters	Define whether the comparison is case-sensitive or case-insensitive during string replacement.
String s is very long, impacting memory usage when generating strings.	Optimize string concatenation to minimize memory allocation overhead and use StringBuilder in Java.
The generated string before reaching n is shorter than K.	Return null or throw an exception because a character at index K doesn't exist in the generated string.
Integer overflow when calculating length of repeated string.	Use appropriate data types (e.g., long) to prevent integer overflow or implement length check prior to string building.

Empty string s or invalid n (n <= 0)

How to Handle:

Return empty string or throw exception, respectively, as no Kth character is possible.

K is out of bounds (K < 0 or K >= length of generated string)

How to Handle:

Return empty string or throw exception since the Kth character does not exist.

n is very large, leading to potentially exponential growth and stack overflow.

How to Handle:

Iterative approach is preferred over recursive to avoid stack overflow issues with large n.

s contains non-alphanumeric characters

How to Handle:

Specify behavior (e.g., strip non-alphanumeric characters or throw an exception if not allowed).

s contains upper and lower case characters

How to Handle:

Define whether the comparison is case-sensitive or case-insensitive during string replacement.

String s is very long, impacting memory usage when generating strings.

How to Handle:

Optimize string concatenation to minimize memory allocation overhead and use StringBuilder in Java.

The generated string before reaching n is shorter than K.

How to Handle:

Return null or throw an exception because a character at index K doesn't exist in the generated string.

Integer overflow when calculating length of repeated string.

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflow or implement length check prior to string building.