[Array] LeetCode 151. Reverse Words in a String

First Solution O(n)

class Solution:
    def reverseWords(self, s: str) -> str:
        result = ""

        n = len(s)
        start, end = n, 0
        count = 0

        for i in range(n-1, -1, -1):
            if s[i] != " " and count == 0:
                end = i
                count = 1
            if s[i] != " " and count == 1:
                start = i
            if s[i] == " " and start <= end and count != 0:
                result += s[start:end+1] + " "
                count = 0

        if s[start:end+1] != " " and count != 0:
                result += s[start:end+1]

        return result if result[-1] != " " else result[:len(result) - 1]

1. Iterate the string from the end.
2. Update the end index to the end index for each word.
3. Update the start index until it meets an empty space in the string s.
4. Add the word to the result if it meets an empty space and if the end index is updated (count != 0).
5. Check once more after iterating for the whole string, if there is a word to add to the result.

Second Solution O(n)

class Solution:
    def reverseWords(self, s: str) -> str:
        res = []
        temp = ""

        for c in s:
            if c != " ":
                temp += c
            elif temp != "":
                res.append(temp)
                temp = ""

        if temp != "":
            res.append(temp)

        return " ".join(res[::-1])

1. Add each character in s to the temp if it is not a blank space.
2. Add the created temp word to result, only if the temp word is not empty.
3. After iterating through the whole string, check once more if the temp is not empty and add it to the result if it is not empty.
4. Join the words in the result in reverse order.