Thomas Ngo

Software Engineer

2 minute read

Problem Statement

problem-151

Intuition

I iterate through the given string in reverse order, identifying words and appending their reversed forms to a result list. This way, I can obtain the reversed words in the correct order.

Approach

I initialize an empty list to store the result and another list to keep track of the characters of the current word. By iterating through the input string in reverse, I identify words and add their reversed forms to the result list. Finally, I will join the reversed words to form the reversed sentence.

Complexity

  • Time complexity: O(n) where n is the length of the input string. The algorithm iterates through the string once.

  • Space complexity: O(n) where n is the length of the input string. The space is used to store the reversed words in the result list.

Code

class Solution:
    def reverseWords(self, s: str) -> str:
        res = []
        N = len(s)
        curr = []
        for i in reversed(range(N)):
            if not s[i].isalnum() and curr:
                res.append(''.join(curr[::-1]))
                curr = []

            if s[i].isalnum():
                curr.append(s[i])

        if curr:
            res.append(''.join(curr[::-1]))
        return ' '.join(res)

Editorial Solution

Approach 1: Built-in Split + Reverse

class Solution:
    def reverseWords(self, s: str) -> str:
        return " ".join(reversed(s.split()))

Approach 2: Reverse the Whole String and Then Reverse Each Word

class Solution:
    def trim_spaces(self, s: str) -> list:
        left, right = 0, len(s) - 1
        # remove leading spaces
        while left <= right and s[left] == ' ':
            left += 1

        # remove trailing spaces
        while left <= right and s[right] == ' ':
            right -= 1

        # reduce multiple spaces to single one
        output = []
        while left <= right:
            if s[left] != ' ':
                output.append(s[left])
            elif output[-1] != ' ':
                output.append(s[left])
            left += 1

        return output

    def reverse(self, l: list, left: int, right: int) -> None:
        while left < right:
            l[left], l[right] = l[right], l[left]
            left, right = left + 1, right - 1

    def reverse_each_word(self, l: list) -> None:
        n = len(l)
        start = end = 0

        while start < n:
            # go to the end of the word
            while end < n and l[end] != ' ':
                end += 1
            # reverse the word
            self.reverse(l, start, end - 1)
            # move to the next word
            start = end + 1
            end += 1

    def reverseWords(self, s: str) -> str:
        # converst string to char array
        # and trim spaces at the same time
        l = self.trim_spaces(s)

        # reverse the whole string
        self.reverse(l, 0, len(l) - 1)

        # reverse each word
        self.reverse_each_word(l)

        return ''.join(l)

Approach 3: Deque of Words

from collections import deque
class Solution:
    def reverseWords(self, s: str) -> str:
        left, right = 0, len(s) - 1
        # remove leading spaces
        while left <= right and s[left] == ' ':
            left += 1

        # remove trailing spaces
        while left <= right and s[right] == ' ':
            right -= 1

        d, word = deque(), []
        # push word by word in front of deque
        while left <= right:
            if s[left] == ' ' and word:
                d.appendleft(''.join(word))
                word = []
            elif s[left] != ' ':
                word.append(s[left])
            left += 1
        d.appendleft(''.join(word))

        return ' '.join(d)

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