<p>We have two special characters:</p>

<ul>
	<li>The first character can be represented by one bit <code>0</code>.</li>
	<li>The second character can be represented by two bits (<code>10</code> or <code>11</code>).</li>
</ul>

<p>Given a binary array <code>bits</code> that ends with <code>0</code>, return <code>true</code> if the last character must be a one-bit character.</p>

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

<pre>
<strong>Input:</strong> bits = [1,0,0]
<strong>Output:</strong> true
<strong>Explanation:</strong> The only way to decode it is two-bit character and one-bit character.
So the last character is one-bit character.
</pre>

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

<pre>
<strong>Input:</strong> bits = [1,1,1,0]
<strong>Output:</strong> false
<strong>Explanation:</strong> The only way to decode it is two-bit character and two-bit character.
So the last character is not one-bit character.
</pre>

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

<ul>
	<li><code>1 &lt;= bits.length &lt;= 1000</code></li>
	<li><code>bits[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution:
    def isOneBitCharacter(self, bits: List[int]) -> bool:
        counter = 0
        while (counter<len(bits)-1):
            print(counter)
            if(bits[counter]==1):
                counter+=2
            
            else:
                counter+=1
        if(counter == len(bits)-1):
            return True
        return False
        
        
16700