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 {
public:
    bool isLongPressedName(string name, string typed) {
        int m = name.length(), n = typed.length();
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (name[i] != typed[j]) {
                return false;
            }
            int x = i + 1;
            while (x < m && name[x] == name[i]) {
                ++x;
            }
            int y = j + 1;
            while (y < n && typed[y] == typed[j]) {
                ++y;
            }
            if (x - i > y - j) {
                return false;
            }
            i = x;
            j = y;
        }
        return i == m && j == n;
    }
};