Long Pressed Name
<p>Your friend is typing his <code>name</code> into a keyboard. Sometimes, when typing a character <code>c</code>, the key might get <em>long pressed</em>, and the character will be typed 1 or more times.</p>

<p>You examine the <code>typed</code> characters of the keyboard. Return <code>True</code> if it is possible that it was your friends name, with some characters (possibly none) being long pressed.</p>

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

<pre>
<strong>Input:</strong> name = &quot;alex&quot;, typed = &quot;aaleex&quot;
<strong>Output:</strong> true
<strong>Explanation: </strong>&#39;a&#39; and &#39;e&#39; in &#39;alex&#39; were long pressed.
</pre>

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

<pre>
<strong>Input:</strong> name = &quot;saeed&quot;, typed = &quot;ssaaedd&quot;
<strong>Output:</strong> false
<strong>Explanation: </strong>&#39;e&#39; must have been pressed twice, but it was not in the typed output.
</pre>

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

<ul>
	<li><code>1 &lt;= name.length, typed.length &lt;= 1000</code></li>
	<li><code>name</code> and <code>typed</code> consist of only lowercase English letters.</li>
</ul>
Two Pointers; String
class Solution:
    def isLongPressedName(self, name: str, typed: str) -> bool:
        m, n = len(name), len(typed)
        i = j = 0
        while i < m and j < n:
            if name[i] != typed[j]:
                return False
            x = i + 1
            while x < m and name[x] == name[i]:
                x += 1
            y = j + 1
            while y < n and typed[y] == typed[j]:
                y += 1
            if x - i > y - j:
                return False
            i, j = x, y
        return i == m and j == n
