<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>

from typing import Set

class Solution:
    # a binary string
    def hasAllCodes(self, s: str, k: int) -> bool:
        # build hash set - all binary_code_hash for length k
        # sliding window of length k, set()
        
        # all_binary_code_hashes: Set[int] = self._build_binary_code_hash(k)

        binary_code_not_seen_count = 1 << k
        binary_code_seen: List[bool] = [False] * binary_code_not_seen_count


        l = r = 0
        hash: int = 0
        while r < len(s):
            hash = hash * 2 + int(s[r])

            # k: 2
            # 101
            # val * (2**k)
            if r - l + 1 > k:
                hash -= int(s[l]) * (2**k)
                l += 1

            if r - l + 1 == k:
                if not binary_code_seen[hash]:
                    binary_code_seen[hash] = True
                    binary_code_not_seen_count -= 1
                    if binary_code_not_seen_count == 0:
                        return True
                

            r += 1

        return False
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