<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:
        size = 1 << k
        found = set()

        if len(s) - k + 1 < size: # not enough substrings of length k
            return False

        cur, i = 0, 0

        while i < k:
            cur <<= 1
            cur += int(s[i])
            i += 1

        found.add(cur)
        mask = (1 << (k - 1)) ^ ((1 << k) - 1)
        # 1 << (k - 1) is simply k bits of 100...0. We want to complement it to get 011...1 and AND it with number to erase the first bit.
        # Since default ~ operator returns two's complement, we will use XOR with 111...1 to complement a number. That is XOR-ing with (1 << k) - 1.

        while i < len(s):
            cur &= mask # remove leftmost bit
            cur <<= 1 # add 0 to the right side
            cur |= int(s[i]) # add 1 if needed
            found.add(cur)
            i += 1

        return len(found) == size
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