<p>Your laptop keyboard is faulty, and whenever you type a character <code>&#39;i&#39;</code> on it, it reverses the string that you have written. Typing other characters works as expected.</p>

<p>You are given a <strong>0-indexed</strong> string <code>s</code>, and you type each character of <code>s</code> using your faulty keyboard.</p>

<p>Return <em>the final string that will be present on your laptop screen.</em></p>

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

<pre>
<strong>Input:</strong> s = &quot;string&quot;
<strong>Output:</strong> &quot;rtsng&quot;
<strong>Explanation:</strong> 
After typing first character, the text on the screen is &quot;s&quot;.
After the second character, the text is &quot;st&quot;. 
After the third character, the text is &quot;str&quot;.
Since the fourth character is an &#39;i&#39;, the text gets reversed and becomes &quot;rts&quot;.
After the fifth character, the text is &quot;rtsn&quot;. 
After the sixth character, the text is &quot;rtsng&quot;. 
Therefore, we return &quot;rtsng&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;poiinter&quot;
<strong>Output:</strong> &quot;ponter&quot;
<strong>Explanation:</strong> 
After the first character, the text on the screen is &quot;p&quot;.
After the second character, the text is &quot;po&quot;. 
Since the third character you type is an &#39;i&#39;, the text gets reversed and becomes &quot;op&quot;. 
Since the fourth character you type is an &#39;i&#39;, the text gets reversed and becomes &quot;po&quot;.
After the fifth character, the text is &quot;pon&quot;.
After the sixth character, the text is &quot;pont&quot;. 
After the seventh character, the text is &quot;ponte&quot;. 
After the eighth character, the text is &quot;ponter&quot;. 
Therefore, we return &quot;ponter&quot;.</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
	<li><code>s[0] != &#39;i&#39;</code></li>
</ul>

class Solution:
    def finalString(self, s: str) -> str:
        brokey_keyboard_arr = []
        for i in range(len(s)):
            if s[i] == 'i':
                brokey_keyboard_arr.reverse()
            else:
                brokey_keyboard_arr.append(s[i])

        brokey_keyboard_s = "".join(brokey_keyboard_arr)
        return brokey_keyboard_s
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