<p>Given a string <code>s</code>, reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;Let&#39;s take LeetCode contest&quot;
<strong>Output:</strong> &quot;s&#39;teL ekat edoCteeL tsetnoc&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;Mr Ding&quot;
<strong>Output:</strong> &quot;rM gniD&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>s</code> contains printable <strong>ASCII</strong> characters.</li>
	<li><code>s</code> does not contain any leading or trailing spaces.</li>
	<li>There is <strong>at least one</strong> word in <code>s</code>.</li>
	<li>All the words in <code>s</code> are separated by a single space.</li>
</ul>

# class Solution:
#     def reverseWords(self, s: str) -> str:
#         s = list(s) # convert to array list
#         l = 0
#         for r in range(len(s)):
#             if s[r] == " " or r == len(s) - 1:
#                 temp_l , temp_r = l , r - l
#                 if r == len(s) - 1:
#                     temp_r = r
#                 while temp_l < temp_r:
#                     s[temp_l], s[temp_r] = s[temp_r], s[temp_l]
#                     temp_l += 1
#                     temp_r -= 1
#                 l = r + 1  # beginning of next word after space
#         return "".join(s)


class Solution:
    def reverseWords(self, s: str) -> str:
        s = list(s)
        l = 0

        for r in range(len(s)):
            if s[r] == " " or r == len(s) - 1:
                temp_l , temp_r = l , r - 1

                if r == len(s) - 1:
                    temp_r = r
                while temp_l < temp_r:
                    s[temp_l] , s[temp_r] = s[temp_r] , s[temp_l]
                    temp_l += 1
                    temp_r -= 1
                l = r + 1
        return "".join(s)                       
17300