<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:
    bool queryString(string s, int n) {
        int len = 0;
        while( ( 1 << len ) <= n )
        {
            len++;
        } 
        //cout << len << endl;
        unordered_set<string> uset;
        for(int i = 0; i < s.size(); i++)
        {
            for(int j = 1; j <= len && i + j <= s.size() ; j++)
            {
                //cout << s.substr(i,j) << endl;
                uset.insert(s.substr(i,j));
            }
        }

        for(int i = 1; i <= n; i++)
        {
            int temp = i;
            string str;
            while( temp > 0 )
            {
                str = "0" + str;
                if( temp % 2 == 1) str[0] = '1';
                temp = temp /2 ;
            }
            if (!uset.count(str)) 
            {
                //cout << i   << " = >"
                //     << str << endl;
                return false;
            }
        }

        return true;
    }
};
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