Jobs

Interviews

Discussions

Reverse String II

Easy

Google

1 view

Topics:

ArraysStringsTwo Pointers

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.

If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.

For example:

s = "abcdefg", k = 2. The expected output is "bacdfeg".
s = "abcd", k = 2. The expected output is "bacd".

Could you provide an algorithm to solve this problem? Consider edge cases such as an empty string or when the string length is not a multiple of k or 2k. What is the time and space complexity of your solution?

Reverse String II

Easy

Google

1 view

Topics:

ArraysStringsTwo Pointers

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.

For example:

s = "abcdefg", k = 2. The expected output is "bacdfeg".
s = "abcd", k = 2. The expected output is "bacd".

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 data type of the input string 's', and what is the maximum possible length of the string?
What is the data type and range of the integer 'k', and what happens if k is zero or negative?
If the length of the string 's' is not a multiple of 'k', how should the remaining characters at the end be handled?
If 'k' is larger than the length of 's', should I reverse the entire string or do nothing?
Is the input string guaranteed to contain only printable ASCII characters, or might it include Unicode characters or null bytes?

Brute Force Solution

Approach

The basic approach to reversing parts of a string is to go through the string and reverse specific segments. For each segment that needs reversing, we'll perform the reversal directly within that portion of the string.

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

Start at the beginning of the string.
Check if there's a segment of the string that needs to be reversed based on the given interval.
If a segment needs reversing, create a temporary copy of just that section.
Reverse the temporary copy of the segment.
Overwrite the original segment of the string with the reversed segment.
Move to the next section of the string based on the given interval and repeat the process of checking for reversal and performing it if needed.
Continue until the entire string has been checked and the appropriate segments have been reversed.

Code Implementation

def reverse_string_ii_brute_force(input_string, interval):    string_length = len(input_string)
    string_as_list = list(input_string)

    for start_index in range(0, string_length, 2 * interval):
        # Determine the end index for the substring to be reversed
        end_index = min(start_index + interval, string_length)

        # Create a copy of the substring to be reversed
        substring_to_reverse = string_as_list[start_index:end_index]

        # Reverse the substring        reversed_substring = substring_to_reverse[::-1]

        # Overwrite the original string with the reversed substring
        string_as_list[start_index:end_index] = reversed_substring

    return "".join(string_as_list)

Big(O) Analysis

Time Complexity

O(n) – The outer loop iterates through the string of length n with a step of k (where k is a constant). Inside the loop, a substring of at most length k is reversed. Reversing a substring of length k takes O(k) time. Since k is constant, the reversal operation takes constant time. Therefore, the overall time complexity is O(n * k) which simplifies to O(n) because k is constant.

Space Complexity

O(k) – The provided solution creates a temporary copy of the string segment that is about to be reversed. In the worst case, this segment can have a length of k. No other data structures that scale with the input string length N are used. Thus, the auxiliary space is dominated by the temporary copy, which has a length of k, leading to a space complexity of O(k).

Optimal Solution

Approach

The best way to reverse parts of a string is to think of it as working in chunks. We'll go through the string section by section, reversing only the needed parts and skipping the rest.

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

First, split the string into chunks of a specific length.
For every other chunk, reverse the order of the characters within that chunk.
Leave the other chunks as they are, without reversing them.
Finally, combine all the chunks back together to form the result.

Code Implementation

def reverse_string_in_chunks(input_string, chunk_size):
    string_length = len(input_string)
    result = ''

    for i in range(0, string_length, chunk_size):
        chunk = input_string[i:i + chunk_size]
        
        # Reverse every other chunk as required
        if (i // chunk_size) % 2 == 1:
            chunk = chunk[::-1]

        result += chunk

    return result

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string of length n in chunks. For each chunk, the algorithm reverses the characters within that chunk only if the chunk is in an 'every other' position. Reversing a chunk takes time proportional to the size of the chunk, which is at most k (where k is the reversal size). Since each character in the original string is visited and possibly reversed at most once, and the chunk size k is considered a constant factor, the overall time complexity is O(n), where n is the length of the string.

Space Complexity

O(N) – The algorithm splits the string into chunks, reverses every other chunk, and combines the chunks. Although not explicitly stated in the plain English explanation, reversing a chunk typically requires creating a temporary copy of that chunk to perform the reversal. In the worst-case scenario where the chunk size is comparable to the string length, the reversed chunks could require temporary storage proportional to the length of the input string. Therefore, the auxiliary space used is proportional to the input size N, where N is the length of the string.

Edge Cases

Case	How to Handle
Null or empty string s	Return an empty string or throw an IllegalArgumentException depending on the requirements.
k is zero	Treat k=0 as no reversal is needed for any section, return the original string.
k is a large number exceeding string length	Treat k as string length; reverse the beginning up to the string length and then continue normal slicing.
String length less than k	Reverse the entire string, as if k was the string length.
String length between k and 2k	Reverse the first k characters and leave the remaining characters as they are.
String contains unicode characters	Ensure the reversal algorithm correctly handles multi-byte unicode characters to prevent corruption.
Very long string to avoid memory issues	Consider in-place reversal or using a StringBuilder for better memory efficiency.
k is negative	Throw an exception or handle it as k = abs(k).

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 data type of the input string 's', and what is the maximum possible length of the string?
What is the data type and range of the integer 'k', and what happens if k is zero or negative?
If the length of the string 's' is not a multiple of 'k', how should the remaining characters at the end be handled?
If 'k' is larger than the length of 's', should I reverse the entire string or do nothing?
Is the input string guaranteed to contain only printable ASCII characters, or might it include Unicode characters or null bytes?

Brute Force Solution

Approach

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

Start at the beginning of the string.
Check if there's a segment of the string that needs to be reversed based on the given interval.
If a segment needs reversing, create a temporary copy of just that section.
Reverse the temporary copy of the segment.
Overwrite the original segment of the string with the reversed segment.
Move to the next section of the string based on the given interval and repeat the process of checking for reversal and performing it if needed.
Continue until the entire string has been checked and the appropriate segments have been reversed.

Code Implementation

def reverse_string_ii_brute_force(input_string, interval):    string_length = len(input_string)
    string_as_list = list(input_string)

    for start_index in range(0, string_length, 2 * interval):
        # Determine the end index for the substring to be reversed
        end_index = min(start_index + interval, string_length)

        # Create a copy of the substring to be reversed
        substring_to_reverse = string_as_list[start_index:end_index]

        # Reverse the substring        reversed_substring = substring_to_reverse[::-1]

        # Overwrite the original string with the reversed substring
        string_as_list[start_index:end_index] = reversed_substring

    return "".join(string_as_list)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

The best way to reverse parts of a string is to think of it as working in chunks. We'll go through the string section by section, reversing only the needed parts and skipping the rest.

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

First, split the string into chunks of a specific length.
For every other chunk, reverse the order of the characters within that chunk.
Leave the other chunks as they are, without reversing them.
Finally, combine all the chunks back together to form the result.

Code Implementation

def reverse_string_in_chunks(input_string, chunk_size):
    string_length = len(input_string)
    result = ''

    for i in range(0, string_length, chunk_size):
        chunk = input_string[i:i + chunk_size]
        
        # Reverse every other chunk as required
        if (i // chunk_size) % 2 == 1:
            chunk = chunk[::-1]

        result += chunk

    return result

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty string s	Return an empty string or throw an IllegalArgumentException depending on the requirements.
k is zero	Treat k=0 as no reversal is needed for any section, return the original string.
k is a large number exceeding string length	Treat k as string length; reverse the beginning up to the string length and then continue normal slicing.
String length less than k	Reverse the entire string, as if k was the string length.
String length between k and 2k	Reverse the first k characters and leave the remaining characters as they are.
String contains unicode characters	Ensure the reversal algorithm correctly handles multi-byte unicode characters to prevent corruption.
Very long string to avoid memory issues	Consider in-place reversal or using a StringBuilder for better memory efficiency.
k is negative	Throw an exception or handle it as k = abs(k).