<p>Given a binary string <code>s</code> and an integer <code>k</code>, return <code>true</code> <em>if every binary code of length</em> <code>k</code> <em>is a substring of</em> <code>s</code>. Otherwise, return <code>false</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;00110110&quot;, k = 2
<strong>Output:</strong> true
<strong>Explanation:</strong> The binary codes of length 2 are &quot;00&quot;, &quot;01&quot;, &quot;10&quot; and &quot;11&quot;. They can be all found as substrings at indices 0, 1, 3 and 2 respectively.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;0110&quot;, k = 1
<strong>Output:</strong> true
<strong>Explanation:</strong> The binary codes of length 1 are &quot;0&quot; and &quot;1&quot;, it is clear that both exist as a substring. 
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;0110&quot;, k = 2
<strong>Output:</strong> false
<strong>Explanation:</strong> The binary code &quot;00&quot; is of length 2 and does not exist in the array.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 5 * 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
	<li><code>1 &lt;= k &lt;= 20</code></li>
</ul>

class Solution:
    def hasAllCodes(self, s: str, k: int) -> bool:
        #Special Case Handle
        #Can be done Better Later
        if len(s)-k+1 < 1<<k:
            return False
        #Enumerate ALL possible substring
        subDict = {}
        for i in range(len(s)-(k-1)):
            sub = str(s[i:i+k])
            if not sub in subDict:
                subDict[sub] = 1
        if len(subDict)==1<<k:
            return True
        else:
            return False 

        
        
        return True
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