<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:
        len_s = len(s)
        if n > len_s * (len(s) + 1) // 2:
            return False
        values = set()
        for index in range(len_s):
            current_value = 0
            for char in s[index : ]:
                current_value *= 2
                if char == '1':
                    current_value += 1
                if current_value <= n:
                    if current_value:
                        values.add(current_value)
                else:
                    break
        return len(values) == n
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