<p>Given a binary string <code>s</code> and a positive integer <code>n</code>, return <code>true</code><em> if the binary representation of all the integers in the range </em><code>[1, n]</code><em> are <strong>substrings</strong> of </em><code>s</code><em>, or </em><code>false</code><em> otherwise</em>.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within a string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "0110", n = 3
<strong>Output:</strong> true
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "0110", n = 4
<strong>Output:</strong> false
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</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;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def queryString(self, s: str, n: int) -> bool:
        nums = set()
        
        if len(s) * min(32, len(s)) < n:
            return False
        
        for length in range(1, min(32, len(s) + 1)):
            p = (1 << length) - 1
            cur = int(s[:length], 2)
            nums.add(cur)
            for i in range(len(s) - length):
                nums.add(cur)
                cur = ((cur << 1) + int(s[i + length])) & p
            nums.add(cur)
        
        for i in range(1, n + 1):
            if i not in nums:
                return False
        
        return True
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