<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 {
public:
    bool backspaceCompare(string s, string t) {
        int i = s.size() - 1, j = t.size() - 1;
        int skipS = 0, skipT = 0;

        while (i >= 0 || j >= 0) {
            while (i >= 0) {
                if (s[i] == '#') {
                    skipS++;
                    i--;
                } else if (skipS > 0) {
                    skipS--;
                    i--;
                } else {
                    break;
                }
            }
            while (j >= 0) {
                if (t[j] == '#') {
                    skipT++;
                    j--;
                } else if (skipT > 0) {
                    skipT--;
                    j--;
                } else {
                    break;
                }
            }
            if (i >= 0 && j >= 0 && s[i] != t[j]) {
                return false;
            }
            if ((i >= 0) != (j >= 0)) {
                return false;
            }
            i--;
            j--;
        }

        return true;
    }
};

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