Reverse Words in a String II

class Solution {
public:
    void reverseStringWords(vector<char>& strVec) {
        reverse(strVec.begin(), strVec.end());
    
        int wordStart = 0, wordEnd = 0;
        int length = strVec.size();
    
        while (wordStart < length) {
            while (wordEnd < length && strVec[wordEnd] != ' ') {
                wordEnd++;
            }
    
            reverse(strVec.begin() + wordStart, strVec.begin() + wordEnd);
    
            wordEnd++;
            wordStart = wordEnd;
        }
    }
};

This solution, aimed at the problem of reversing the words in a string using C++, employs an efficient approach to manipulate string data directly in place for enhanced performance and reduced memory usage.

The solution involves the following steps:

1. Reverse the entire string initially to invert the order of characters.
2. Iterate over the reversed string to identify individual words separated by spaces.
3. Reverse each word found to restore its correct order while keeping the overall reversed word order intact.

Key details of the implementation:

• Utilize the reverse function from the C++ Standard Library for in-place string manipulation.
• Identify the start and end of each word using a loop, and apply the reverse function to each word independently.

By reversing the entire string first and then each word, the solution achieves the goal of reversing the order of words in the string while maintaining the order of characters within each word. This method avoids the need for additional storage, thus being memory efficient.

class Solution {
    public void reverseSegment(char[] s, int startIdx, int endIdx) {
        while (startIdx < endIdx) {
            char temporary = s[startIdx];
            s[startIdx++] = s[endIdx];
            s[endIdx--] = temporary;
        }
    }
    
    public void reverseEachWordInArray(char[] s) {
        int length = s.length;
        int startPos = 0, endPos = 0;
    
        while (startPos < length) {
            // find end of the current word
            while (endPos < length && s[endPos] != ' ') ++endPos;
            // reverse current word
            reverseSegment(s, startPos, endPos - 1);
            // adjust to the start of next word
            startPos = endPos + 1;
            ++endPos;
        }
    }
    
    public void reverseWordsInSentence(char[] s) {
        // reverse the entire character array
        reverseSegment(s, 0, s.length - 1);
    
        // then reverse each word in the array
        reverseEachWordInArray(s);
    }
}

This Java solution for reversing the words in a string in-place efficiently manipulates character arrays. Follow these key steps to understand the approach:

• Define a helper method reverseSegment to reverse a segment within a character array between two indices. This method swaps characters from start to end until the entire segment is reversed.

• Implement reverseEachWordInArray to handle the reversal of each word individually:

  • Loop through the character array to identify individual words. Words are segments that end with a space character.
  • Apply the reverseSegment method to each word identified.

• The main method reverseWordsInSentence performs the overall task in two major steps:

  • First, reverse the entire character array to invert the order of all characters, effectively placing all words in their final reversed position in the array.
  • Then, use reverseEachWordInArray to correct the order of characters within each word in the array.

This approach ensures that words are reversed in their position without any additional memory for storage, making the operation space-efficient.

class Solution:
    def reverse_segment(self, arr: List[str], start: int, end: int) -> None:
        while start < end:
            arr[start], arr[end] = arr[end], arr[start]
            start += 1
            end -= 1
    
    def reverse_word_by_word(self, arr: List[str]) -> None:
        length = len(arr)
        begin = finish = 0
    
        while begin < length:
            while finish < length and arr[finish] != ' ':
                finish += 1
            self.reverse_segment(arr, begin, finish - 1)
            begin = finish + 1
            finish += 1
                
    def reverseWords(self, s: List[str]) -> None:
        self.reverse_segment(s, 0, len(s) - 1)
        self.reverse_word_by_word(s)

The provided Python solution focuses on reversing the words in a given list of characters, effectively solving the "Reverse Words in a String II" problem without returning any value, as it modifies the list in place. This operation is done by defining and utilizing two main helper functions: reverse_segment and reverse_word_by_word.

Here's how the code achieves its goal:

• Initial Full Reverse: *Applies the reverse_segment function initially to the entire string. This action reverses the whole array of characters. For example, given "the sky is blue", it transforms the list to "eulb si yks eht".

• Word-by-Word Reverse:

  • Executes the reverse_word_by_word method, which then processes each word individually. It scans for word boundaries (spaces) and applies reverse_segment to each word segment to restore the original word order while keeping the individual words reversed.

Here's what happens during the execution:

1. reverse_segment: Takes a segment of the array defined by start and end indices, and reverses the characters within that segment in place using a while loop.
2. reverse_word_by_word: Iterates through the complete list of characters. For each word detected (delimited by spaces), it applies reverse_segment to reverse back each word in its place.
3. reverseWords: Combines these functions methodically. It reverses the whole character list first to achieve the reversed word order. Then, it reverses each word to correct the orientation of the characters within words.

This method is both time-efficient and space-efficient as it optimizes in-place operations without utilizing additional memory for another array, fitting well within typical constraints for string manipulation challenges.