Backspace String Compare
<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>
Stack; Two Pointers; String; Simulation
class Solution:
    def backspaceCompare(self, s: str, t: str) -> bool:
        i, j, skip1, skip2 = len(s) - 1, len(t) - 1, 0, 0
        while i >= 0 or j >= 0:
            while i >= 0:
                if s[i] == '#':
                    skip1 += 1
                    i -= 1
                elif skip1:
                    skip1 -= 1
                    i -= 1
                else:
                    break
            while j >= 0:
                if t[j] == '#':
                    skip2 += 1
                    j -= 1
                elif skip2:
                    skip2 -= 1
                    j -= 1
                else:
                    break
            if i >= 0 and j >= 0:
                if s[i] != t[j]:
                    return False
            elif i >= 0 or j >= 0:
                return False
            i, j = i - 1, j - 1
        return True
