<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:
    long long conv(string x){
        int sz = x.size();
        long long p=1;
        long long ans = 0;
        for(int i=sz-1; i>=0; i--){
            if(x[i] == '1')ans += p;
            p *= 2;
        }
        return ans;
    }
bool queryString(string s, int n) {
int sz = s.size();
        unordered_map<long long, bool> p;
      //  unsigned long long sum = n*(n+1)/2;
        for(int i=0; i<sz; i++){
        //   if(sum < 0)return false;
            if(s[i] == '0')continue;
            string ss = "";
            for(int j=i; j<sz; j++){
                ss += s[j];
                if(j-i+1 > 31)break;
                long long ha = conv(ss);
                if(ha <= n && !p[ha]){
                   // sum -= ha;
                   /// cout<<ha<<endl;
                    p[ha] = true;
                }
            }
        }
        for(int i=1; i<=n; i++){
            if(!p[i])return false;
        }
        return true;
    }

};
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