<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>

class Solution:
    def isLongPressedName(self, name: str, typed: str) -> bool:
        i, j = 0, 0
        while i < len(name) and j < len(typed):
            count_ch_named, count_ch_typed = 1, 1

            if name[i] != typed[j]:
                return False
        
            # Occurrences of ch in name
            while i < len(name) - 1  and name[i+1] == name[i]:
                i += 1
                count_ch_named += 1

            # Occurrences of ch in typed
            while j < len(typed) - 1   and typed[j+1] == typed[j]:
                j += 1
                count_ch_typed += 1

            if count_ch_named > count_ch_typed:
                return False

            i += 1
            j += 1

        if i != len(name) or j != len(typed):
            return False
        return True

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