<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:
    int getVal(string s){
        int val=0;
        for(int i=0;i<s.size();i++){
            val=val*2+(s[i]-'0');
        }
        return val;
    }
    bool queryString(string s, int n) {
        vector<int> x;
        int N=s.size();
        for(int i=0;i<N;i++){
            long long int val=0;
            for(int j=i;j<N;j++){
                if(val>1e9+1)   break;
                val=val*2+(s[j]-'0');
                x.push_back(val);
            }
        }
        sort(x.begin(),x.end());
        int i=0,val=1;
        bool cond=false;
        while(i<x.size()){
            int currVal=x[i];
            while(i<x.size()&&currVal==x[i])
                i++;
            if(currVal){
                cond=true;
                if(val>n)   return true;
                if(currVal>val) return false;
                val++;
            }
        }
        
        return cond;
    }
};
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