// Time:  O(n^2), n is the length of S
// Space: O(1)
    
class Solution {
public:
    bool queryString(string S, int N) {
        // since S with length n has at most different n-k+1 k-digit numbers
        // => given S with length n, valid N is at most 2(n-k+1)
        // => valid N <= 2(n-k+1) < 2n = 2 * S.length
        for (int i = N; i > N / 2; --i) {
            // if bin_i is a substring of S, bin_i/2 is too
            const auto& b = bitset<32>(i).to_string();
            if (S.find(b.substr(b.find("1"))) == string::npos) {
                return false;
            }
        }
        return true;
    }
};