algoadvance

Given a string s and an integer k, you need to perform the following action on the string:

• 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.

Example 1:

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

Example 2:

Input: s = "abcd", k = 2
Output: "bacd"

Clarifying Questions

1. Q: Can s contain any type of character (e.g., numbers, punctuation)? A: Yes, s can contain any type of character.

2. Q: Is the given string s guaranteed to be non-empty? A: Typically, Yes, but if an empty string is provided, the output should also be an empty string.

3. Q: Can the integer k be greater than the length of the string? A: Yes, the integer k might be greater than the length of the string, and it should handle this situation gracefully.

Strategy

1. Iterate over the string in segments of 2k.
2. For each segment, first reverse the first k characters.
3. If fewer than k characters are left in any remaining segment, reverse all of those characters.

Code

def reverseStr(s: str, k: int) -> str:
    s = list(s)  # Convert to list for mutability
    for i in range(0, len(s), 2 * k):
        s[i:i+k] = reversed(s[i:i+k])
    return "".join(s)

Time Complexity

• Time Complexity: The code loops through the string in chunks of size 2k, performing a reversal operation on fixed-sized chunks. Given that slicing and reversing subarrays are linear with respect to the chunk size, the overall time complexity is O(n), where n is the length of the string s.

• Space Complexity: The space complexity is O(n), since in order to handle in-place modifications and the final join operation, we may use extra space proportional to the input size.

This solution ensures each segment is handled correctly according to the given rules and efficiently processes the string in linear time.

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