LeetCode 557: Reverse Words in a String III Solution in Python – A Step-by-Step Guide

What Is LeetCode 557: Reverse Words in a String III?

In LeetCode 557: Reverse Words in a String III, you’re given a string s containing words separated by single spaces, and your task is to return a new string where each word’s characters are reversed, while the word order and spaces remain unchanged. For example, with s = "Let's take LeetCode contest", the result is "s'teL ekat edoCteeL tsetnoc". This problem builds on LeetCode 344: Reverse String for basic reversal but adds the complexity of handling multiple words within a single string.

Problem Statement

• Input: s (str)—string with words separated by single spaces.
• Output: String—each word reversed, spaces and order preserved.
• Rules: Words are sequences of characters separated by single spaces; reverse each word’s characters.

Constraints

• 1 <= s.length <= 5 * 10⁴
• s contains printable ASCII characters.
• No leading or trailing spaces.
• At least one word, separated by single spaces.

Examples

• Input: s = "Let's take LeetCode contest"
• Output: "s'teL ekat edoCteeL tsetnoc"
• Each word reversed: "Let's" → "s'teL", "take" → "ekat", etc.
• Input: s = "God Ding"
• Output: "doG gniD"
• "God" → "doG", "Ding" → "gniD".
• Input: s = "hello"
• Output: "olleh"
• Single word reversed.

Understanding the Problem: Flipping Words

To solve LeetCode 557: Reverse Words in a String III in Python, we need to reverse the characters of each word in a string while keeping the spaces and word order intact. A naive approach might manually find word boundaries and reverse each, but with lengths up to 5 * 10⁴, we can optimize by either processing in-place or using Python’s built-in tools efficiently. The problem tests string traversal and manipulation skills. We’ll explore:

• Best Solution (Two-Pointer Word Reversal): O(n) time, O(1) space—fast and space-efficient (excluding output).
• Alternative Solution (Split and Join with Reversal): O(n) time, O(n) space—simple but uses extra memory.

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

Best Solution: Two-Pointer Word Reversal

Why Two-Pointer Wins

The two-pointer word reversal is the best for LeetCode 557 because it reverses each word in-place (after converting to a list) in O(n) time and O(1) extra space (excluding the mutable list required by Python strings’ immutability), efficiently traversing the string once to identify and flip words. It’s like walking through the sentence with a marker, flipping each word as you go without needing a big scratchpad!

How It Works

Think of this as a word-flipping dance:

• Step 1: Convert to List:
• Turn string into a list of characters (Python strings are immutable).
• Step 2: Find Word Boundaries:
• Use a pointer to track word start, move until space or end.
• Step 3: Reverse Each Word:
• Use two pointers (start, end) to swap characters in-place.
• Step 4: Move to Next Word:
• Skip space, repeat until end.
• Step 5: Join and Return:
• Convert list back to string.
• Why It Works:
• In-place reversal minimizes space.
• Single pass keeps time linear.

It’s like a string-word twirler!

Step-by-Step Example

Example: s = "Let's take"

• Step 1: Convert: chars = ['L', 'e', 't', "'", 's', ' ', 't', 'a', 'k', 'e'].
• Step 2: First word (0 to 4):
• Start = 0, End = 4 (before space).
• Reverse: ['s', "'", 't', 'e', 'L', ' ', 't', 'a', 'k', 'e'].
• Step 3: Next word (6 to 9):
• Start = 6, End = 9.
• Reverse: ['s', "'", 't', 'e', 'L', ' ', 'e', 'k', 'a', 't'].
• Step 4: Join: "s'teL ekat".
• Result: "s'teL ekat".

Example: s = "God"

• Step 1: chars = ['G', 'o', 'd'].
• Step 2: Reverse (0 to 2):
• ['d', 'o', 'G'].
• Step 3: Join: "doG".
• Result: "doG".

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

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

        # Step 2: Initialize pointers
        start = 0

        # Step 3: Process each word
        for i in range(n + 1):
            # When we hit a space or end of string
            if i == n or chars[i] == ' ':
                # Reverse word from start to i-1
                left, right = start, i - 1
                while left < right:
                    chars[left], chars[right] = chars[right], chars[left]
                    left += 1
                    right -= 1
                # Move start to next word (after space)
                start = i + 1

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

• Line 4: Convert string to mutable list.
• Line 7: Start pointer at first word.
• Lines 10-19: Iterate:
• Detect word end (space or string end).
• Reverse word using two pointers.
• Update start to next word.
• Line 22: Join list back to string.
• Time Complexity: O(n)—single pass with reversals.
• Space Complexity: O(1)—extra space (excluding input list).

It’s like a word-flipping maestro!

Alternative Solution: Split and Join with Reversal

Why an Alternative Approach?

The split and join with reversal solution splits the string into words, reverses each, and joins them back, running in O(n) time and O(n) space due to the word list. It’s simple and leverages Python’s built-in methods, making it a good alternative for readability or when space isn’t a concern.

How It Works

Picture this as a word-chopping flipper:

• Step 1: Split string into words.
• Step 2: Reverse each word’s characters.
• Step 3: Join words with spaces.
• Step 4: Return result.

It’s like a word-reversal assembler!

Step-by-Step Example

Example: s = "God Ding"

• Step 1: Split: ["God", "Ding"].
• Step 2: Reverse: ["doG", "gniD"].
• Step 3: Join: "doG gniD".
• Result: "doG gniD".

Code for Split Approach

Copy codeclass Solution:
    def reverseWords(self, s: str) -> str:
        # Step 1: Split into words
        words = s.split()

        # Step 2: Reverse each word
        reversed_words = [word[::-1] for word in words]

        # Step 3: Join back with spaces
        return ' '.join(reversed_words)

• Line 4: Split on spaces.
• Line 7: List comprehension reverses each word.
• Line 10: Join with single spaces.
• Time Complexity: O(n)—split, reverse, join.
• Space Complexity: O(n)—word list storage.

It’s a split-word flipper!

Comparing the Two Solutions

• Two-Pointer (Best):
• Pros: O(n), O(1), in-place.
• Cons: Slightly more code.
• Split and Join (Alternative):
• Pros: O(n), O(n), concise.
• Cons: Extra space.

Two-pointer wins for efficiency!

Additional Examples and Edge Cases

• "a": "a".
• "hello world": "olleh dlrow".
• " ": "" (invalid per constraints, assume handled).

Two-pointer handles them all!

Complexity Recap

• Two-Pointer: Time O(n), Space O(1).
• Split and Join: Time O(n), Space O(n).

Two-pointer’s the space champ!

Key Takeaways

• Two-Pointer: In-place flipping—learn at Python Basics!
• Split: Simple slicing.
• Strings: Words are fun.
• Python: Pointers or split, your pick!

Final Thoughts: Flip Those Words!

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