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
func isLongPressedName(name string, typed string) bool {
	m, n := len(name), len(typed)
	i, j := 0, 0

	for i < m && j < n {
		if name[i] != typed[j] {
			return false
		}
		x, y := i+1, j+1

		for x < m && name[x] == name[i] {
			x++
		}

		for y < n && typed[y] == typed[j] {
			y++
		}

		if x-i > y-j {
			return false
		}

		i, j = x, y
	}

	return i == m && j == n
}