LeetCode 541: Reverse String II Solution in Python – A Step-by-Step Guide

What Is LeetCode 541: Reverse String II?

In LeetCode 541: Reverse String II, you’re given a string s and an integer k, and your task is to reverse the first k characters for every 2k characters in the string, leaving any remaining characters (less than k) unchanged. For example, with s = "abcdefg" and k = 2, you reverse "ab" to "ba," "cd" to "dc," and leave "efg" as is, yielding "bacdfeg." This problem builds on LeetCode 344: Reverse String for basic string reversal but adds the twist of segmented processing.

Problem Statement

• Input:
• s (str): Input string.
• k (int): Number of characters to reverse per segment.
• Output: String—modified string with every k characters reversed in 2k segments.
• Rules: Reverse first k chars in each 2k segment; if fewer than k chars remain, leave as is.

Constraints

• 1 <= s.length <= 10⁴
• s contains only lowercase English letters.
• 1 <= k <= 10⁴

Examples

• Input: s = "abcdefg", k = 2
• Output: "bacdfeg"
• Segments: "ab" → "ba," "cd" → "dc," "efg" unchanged.
• Input: s = "abcd", k = 2
• Output: "bacd"
• "ab" → "ba," "cd" → "cd" (only 2 chars, reverse first k).
• Input: s = "abcdefg", k = 8
• Output: "gfedcba"
• One segment of 7 < 8, reverse all.

Understanding the Problem: Flipping Chunks

To solve LeetCode 541: Reverse String II in Python, we need to reverse the first k characters in every 2k segment of a string, leaving any trailing characters (less than k) unchanged. The string is processed in blocks of 2k, and within each, only the first k chars flip. A naive approach might reverse the entire string and adjust, but with lengths up to 10⁴, we can optimize by processing segments directly. We’ll explore:

• Best Solution (Two-Pointer with Step-Wise Reversal): O(n) time, O(1) space—efficient and in-place.
• Alternative Solution (String Slicing with List Conversion): O(n) time, O(n) space—clear but uses extra memory.

Let’s dive into the two-pointer solution with a friendly breakdown!

Best Solution: Two-Pointer with Step-Wise Reversal

Why Two-Pointer Wins

The two-pointer with step-wise reversal is the best for LeetCode 541 because it processes the string in-place in O(n) time and O(1) space, reversing each k-sized chunk within 2k segments using two pointers to swap characters. It’s like flipping cards in a deck, segment by segment, without needing extra space!

How It Works

Think of this as a string-flipping dance:

• Step 1: Process in 2k Segments:
• Iterate string in steps of 2k.
• Step 2: Reverse k Chars:
• For each segment, use two pointers (left, right) to swap first k chars.
• Adjust right if fewer than k chars remain.
• Step 3: Move to Next Segment:
• Skip to next 2k block.
• Why It Works:
• In-place swaps keep space O(1).
• Step-wise ensures correct k-sized reversals.

It’s like a string-chunk twirler!

Step-by-Step Example

Example: s = "abcdefg", k = 2

• Init: s = list("abcdefg") (for mutability), n = 7.
• Step 1: First 2k = 4 segment (0 to 3):
• Left = 0, Right = min(0 + 2 - 1, 6) = 1.
• Swap s[0] (a) and s[1] (b): "bacdefg".
• Step 2: Next 2k segment (4 to 7):
• Left = 4, Right = min(4 + 2 - 1, 6) = 5.
• Swap s[4] (e) and s[5] (f): "bacdfeg".
• Step 3: Remaining < k (6 to 6), unchanged.
• Result: "bacdfeg".

Example: s = "abcd", k = 2

• Step 1: 2k = 4 segment (0 to 3):
• Left = 0, Right = 1.
• Swap s[0] (a) and s[1] (b): "bacd".
• Step 2: No more 2k, done.
• Result: "bacd".

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def reverseStr(self, s: str, k: int) -> str:
        # Step 1: Convert string to list for mutability
        s_list = list(s)
        n = len(s)

        # Step 2: Process in 2k segments
        for i in range(0, n, 2 * k):
            left = i
            # Right pointer: min of k-1 from left or end of string
            right = min(i + k - 1, n - 1)

            # Step 3: Reverse k chars using two pointers
            while left < right:
                s_list[left], s_list[right] = s_list[right], s_list[left]
                left += 1
                right -= 1

        # Step 4: Convert back to string
        return ''.join(s_list)

• Line 4: Convert string to list (strings are immutable in Python).
• Line 7: Iterate in 2k steps.
• Lines 9-10: Set left to segment start, right to min of k-1 ahead or end.
• Lines 13-16: Swap characters from left to right within k.
• Line 19: Join list back to string.
• Time Complexity: O(n)—single pass with swaps.
• Space Complexity: O(1)—in-place (excluding input list).

It’s like a string-flipping maestro!

Alternative Solution: String Slicing with List Conversion

Why an Alternative Approach?

The string slicing with list conversion solution breaks the string into segments, reverses k-sized chunks using slicing, and recombines them, running in O(n) time but using O(n) space for the list. It’s clear and leverages Python’s slicing, making it great for beginners who prefer readability over space efficiency.

How It Works

Picture this as a string-chunk cutter:

• Step 1: Convert string to list.
• Step 2: Slice into 2k segments.
• Step 3: Reverse first k chars in each, keep rest.
• Step 4: Join back to string.

It’s like a string-chunk slicer!

Step-by-Step Example

Example: s = "abcdefg", k = 2

• Step 1: s_list = list("abcdefg").
• Step 2: Segments: "abcd" (0-4), "efg" (4-7).
• Step 3: Process:
• "abcd": Reverse 0-2 → "ba," append "cd" → "bacd".
• "efg": Reverse 4-6 → "fe," append "g" → "feg".
• Step 4: Join: "bacd" + "feg" = "bacdfeg".
• Result: "bacdfeg".

Code for Slicing Approach

Copy codeclass Solution:
    def reverseStr(self, s: str, k: int) -> str:
        # Step 1: Convert to list
        s_list = list(s)
        n = len(s)

        # Step 2: Process in 2k segments
        for i in range(0, n, 2 * k):
            # Step 3: Slice and reverse k chars
            segment = s_list[i:i + 2 * k]
            k_chars = segment[:k]
            k_chars.reverse()
            s_list[i:i + k] = k_chars

        # Step 4: Join back to string
        return ''.join(s_list)

• Line 4: Convert to list.
• Line 7: Step by 2k.
• Lines 9-12: Slice 2k segment, reverse first k, replace in list.
• Line 15: Join back.
• Time Complexity: O(n)—single pass with reversals.
• Space Complexity: O(n)—list storage.

It’s a slicing string flipper!

Comparing the Two Solutions

• Two-Pointer (Best):
• Pros: O(n), O(1), in-place.
• Cons: Pointer logic.
• Slicing (Alternative):
• Pros: O(n), O(n), readable.
• Cons: Extra space.

Two-pointer wins for efficiency!

Additional Examples and Edge Cases

• "a", k = 2: "a".
• "abcdef", k = 3: "cbadef".
• "abc", k = 1: "abc".

Two-pointer handles them all!

Complexity Recap

• Two-Pointer: Time O(n), Space O(1).
• Slicing: Time O(n), Space O(n).

Two-pointer’s the space champ!

Key Takeaways

• Two-Pointer: In-place wins—learn at Python Basics!
• Slicing: Clear but costly.
• Strings: Chunks are fun.
• Python: Pointers or slices, your pick!

Final Thoughts: Flip Those Chunks!

LeetCode 541: Reverse String II in Python is a delightful string challenge. Two-pointer with step-wise reversal is your fast track, while slicing offers a clear alternative. Want more? Try LeetCode 344: Reverse String or LeetCode 557: Reverse Words in a String III. Ready to reverse? Head to Solve LeetCode 541 on LeetCode and twist that string today!