<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 {
public:
    std::string to_bstring(int n) {
        std::string s;

        while (n > 0) {
            s = ((n ^ (n >> 1 << 1)) ? "1" : "0") + s;
            n = n >> 1;
        }

        return s;
    }
    
    bool queryString(string s, int n) {
        if (n == 0) {return s.find("0") != std::string::npos;}
        int const length = floor(log2(n)) + 1;
        auto const target = to_bstring(n);
        for (int i = 0; i <= s.size() - length; ++i) {
            auto const substr = s.substr(i, length);
            if (substr == target) {
                std::cout << "Found target: " << substr << " n: " << n << std::endl;
                if (n > 1) {
                    return queryString(s, n - 1);
                }
                return true;
            }
        }
        return false;
    }
};
10000