<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:
        if name == typed:
            return True
        s1 = [''.join(g) for _, g in groupby(name)]
        s2 = [''.join(g) for _, g in groupby(typed)]
        if len(s1) != len(s2):
            return False
        for i in range(len(s1)):
            print(s1[i],s2[i],len(s1[i]),len(s2[i]))
            if len(s1[i]) > len(s2[i]) or s1[i][0] != s2[i][0]:
                return False
        return True
            
            
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