<p>Given two strings <code>s</code> and <code>t</code>, return <code>true</code> <em>if they are equal when both are typed into empty text editors</em>. <code>&#39;#&#39;</code> means a backspace character.</p>

<p>Note that after backspacing an empty text, the text will continue empty.</p>

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

<pre>
<strong>Input:</strong> s = &quot;ab#c&quot;, t = &quot;ad#c&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Both s and t become &quot;ac&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;ab##&quot;, t = &quot;c#d#&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Both s and t become &quot;&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a#c&quot;, t = &quot;b&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> s becomes &quot;c&quot; while t becomes &quot;b&quot;.
</pre>

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

<ul>
	<li><code><span>1 &lt;= s.length, t.length &lt;= 200</span></code></li>
	<li><span><code>s</code> and <code>t</code> only contain lowercase letters and <code>&#39;#&#39;</code> characters.</span></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Can you solve it in <code>O(n)</code> time and <code>O(1)</code> space?</p>

class Solution:
    def backspaceCompare(self, s: str, t: str) -> bool:
        s_stack = []
        t_stack = []

        for letter in s:
            if letter != "#":
                s_stack.append(letter)
            else:
                if len(s_stack)>0:
                    s_stack.pop()
        
        for letter in t:
            if letter != "#":
                t_stack.append(letter)
            else:
                if len(t_stack)>0:
                    t_stack.pop()
        
        return s_stack == t_stack
16300