Problem Description

Given a string s that contains words separated by spaces, you need to reverse the characters within each individual word while keeping the words in their original positions and preserving all whitespace.

For example, if the input string is "Let's take LeetCode contest", each word should have its characters reversed individually:

• "Let's" becomes "s'teL"
• "take" becomes "ekat"
• "LeetCode" becomes "edoCteeL"
• "contest" becomes "tsetnoc"

The final output would be "s'teL ekat edoCteeL tsetnoc".

The key requirements are:

1. Reverse the characters in each word
2. Keep the words in their original order
3. Preserve the original spacing between words

The solution splits the string by spaces using s.split(), which creates a list of words. Then for each word t, it reverses the characters using Python's slice notation t[::-1]. Finally, it joins all the reversed words back together with spaces using " ".join().

Intuition

When we look at this problem, we need to perform an operation on each word independently while maintaining the overall structure of the sentence. This naturally suggests breaking the problem down into smaller parts.

The first observation is that we need to process each word separately. Since words are separated by spaces, we can use the space character as a delimiter to identify individual words. This leads us to think about splitting the string into a list of words.

Once we have individual words, the task becomes straightforward - reverse each word. In Python, reversing a string can be elegantly done using slice notation [::-1], which reads the string backwards.

After reversing each word, we need to reconstruct the sentence. Since we want to maintain the original spacing (single spaces between words), we can join the reversed words back together with spaces.

The beauty of this approach is that split() without arguments automatically handles multiple spaces by treating consecutive spaces as a single delimiter and returns only the non-empty words. When we join them back with a single space, we get the desired output format.

This leads to a clean one-liner solution: split the string into words, reverse each word, and join them back. The entire process maintains the word order since we process them in sequence and only modify the internal character order of each word.

Learn more about Two Pointers patterns.

Solution Approach

The implementation follows a straightforward simulation approach where we process the string in three steps:

1. Split the string into words: We use s.split() to break the input string into a list of words. The split() method without arguments automatically splits by any whitespace and removes empty strings from the result.

2. Reverse each word: For each word t in the list, we apply Python's slice notation t[::-1] to reverse it. The slice [::-1] creates a new string that starts from the end and moves backwards with a step of -1, effectively reversing the string.

3. Join the reversed words: We use " ".join() to concatenate all reversed words back into a single string with single spaces between them.

The complete solution combines these steps into a single line using a generator expression:

return " ".join(t[::-1] for t in s.split())

Here's how it works step by step:

• s.split() produces a list like ["Let's", "take", "LeetCode", "contest"]
• The generator expression t[::-1] for t in s.split() iterates through each word and reverses it
• This produces reversed words: "s'teL", "ekat", "edoCteeL", "tsetnoc"
• " ".join() combines these reversed words with spaces: "s'teL ekat edoCteeL tsetnoc"

The time complexity is O(n) where n is the length of the string, as we need to process each character once. The space complexity is also O(n) for storing the split words and the final result string.

Example Walkthrough

Let's walk through a small example with the input string "hello world".

Step 1: Split the string into words

• Input: "hello world"
• After s.split(): ["hello", "world"]
• We now have a list containing two separate words

Step 2: Reverse each word individually

• Process first word: "hello"[::-1] → "olleh"
  • The slice [::-1] reads from the last character 'o' backwards to the first 'h'
• Process second word: "world"[::-1] → "dlrow"
  • Similarly, 'd' is the last character and we read backwards to 'w'
• After reversing all words: ["olleh", "dlrow"]

Step 3: Join the reversed words with spaces

• Use " ".join(["olleh", "dlrow"])
• Result: "olleh dlrow"

The complete process transforms "hello world" into "olleh dlrow", where each word has been reversed in place while maintaining the original word order and spacing structure.

For a more complex example with multiple spaces: if the input were "hello world" (two spaces), split() would still produce ["hello", "world"], and the final output would be "olleh dlrow" with a single space, as split() normalizes the spacing by default.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. This is because:

• s.split() traverses the entire string once to split it into words: O(n)
• For each word t, t[::-1] reverses it, which takes time proportional to the word's length
• The total length of all words combined is at most n (excluding spaces), so reversing all words takes O(n) in total
• " ".join() combines the reversed words, traversing through all characters once: O(n)
• Overall: O(n) + O(n) + O(n) = O(n)

The space complexity is O(n). This is because:

• s.split() creates a list of words that collectively contain up to n characters: O(n)
• The list comprehension creates reversed versions of each word, requiring space for all reversed characters: O(n)
• " ".join() creates a new string of length approximately n: O(n)
• Overall: O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Handling Multiple or Irregular Spaces

The current solution using s.split() without arguments treats multiple consecutive spaces as a single delimiter and removes them entirely. This means if the input has multiple spaces between words or leading/trailing spaces, they won't be preserved in the output.

Example where this fails:

• Input: "Let's take LeetCode" (multiple spaces between words)
• Current output: "s'teL ekat edoCteeL" (single spaces)
• Expected (if preserving spaces): "s'teL ekat edoCteeL"

Solution: If you need to preserve the exact spacing, use s.split(' ') with a single space argument, but then handle empty strings:

def reverseWords(self, s: str) -> str:
    words = s.split(' ')
    return ' '.join(word[::-1] if word else '' for word in words)

Pitfall 2: Memory Usage with Large Strings

The current approach creates multiple intermediate strings (split list, reversed words, final joined string), which can be memory-intensive for very large inputs.

Solution: For memory-critical applications, consider an in-place approach using a list of characters:

def reverseWords(self, s: str) -> str:
    chars = list(s)
    start = 0
  
    for end in range(len(chars) + 1):
        if end == len(chars) or chars[end] == ' ':
            # Reverse the word in place
            left, right = start, end - 1
            while left < right:
                chars[left], chars[right] = chars[right], chars[left]
                left += 1
                right -= 1
            start = end + 1
  
    return ''.join(chars)

Pitfall 3: Unicode and Special Characters

The slice reversal [::-1] works at the character level, which might produce incorrect results for certain Unicode combinations like emojis with modifiers or combining characters.

Example where this might fail:

• Input with emoji: "Hello 👨‍👩‍👧‍👦 World" (family emoji with zero-width joiners)
• The reversal might break the emoji sequence

Solution: For proper Unicode handling, consider using libraries that understand grapheme clusters or implement proper Unicode-aware reversal.