<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:
    string convert(int n)
    {
        string s;
        for(int i=31;i>=0;i--)
        {
            if(1<<i & n)
            {
                s.push_back('1');
            }
            else
            {
                s.push_back('0');
            }
        }
        string temp = s;
        reverse(temp.begin(), temp.end());
        // cout<<"rhhr " <<temp<<endl;
        int ind = temp.size()-1;
        while(temp.back() == '0')
        {
            temp.pop_back();
            // cout<<temp<<endl;
        }
        reverse(temp.begin(), temp.end());
        return temp;
    }
    bool queryString(string s, int n) {
        int m = s.size();
        set<string> v;
        if(n > 2100)
        {
            return false;
        }
        for(int i=0;i<m;i++)
        {
            for(int j=i;j<min(m, i+31);j++)
            {
                string temp = s.substr(i, j-i+1);
                v.insert(temp);
            }
        }
        for(int i=1;i<=n;i++)
        {
            int num = i;
            string s = convert(num);
            if(v.find(s) == v.end())
            {
                return false;
            }
        }
        return true;

    }
};
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