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  2. Algorithms
  3. Strings
  4. Two Two
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Two Two

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  • saintmyjava
     + 0 comments

    Solution worked to 7 test. Java 8

    static HashSet<String> set;

    static {
        set = new HashSet<>();
        int n = 800;
        set.add("1");
        for (int i = 0; i <= n - 1; ++i) {
            set.add(BigInteger.ONE.shiftLeft(i).toString());
        }
    }

    static int twoTwo(String s){
        int count = 0;
        for (int i = 0; i < s.length();i++){
            if (s.charAt(i)=='0') continue;
            for (int j = i; j < s.length(); j++) {
                if (set.contains(s.substring(i,j+1))){
                    count++;
                }
            }
        }
        System.out.println(count);
        return count;
    }

  • yashparihar729
     + 1 comment

    Here is my solution in java, javascript, python, C, C++, csharp HackerRank Two Two Problem Solution

  • vgiargia
     + 0 comments

    the key is to use a trie data structure I did it in C so I need to code a very basic trie DS from scratch

  • mineman1012221
     + 0 comments

    here is the solution of Two Two Click Here

  • ammaralii
     + 1 comment

    I was able to solve it using Python 3 after 7 attempts and got top 34% so I optimized it further using bit operator but same result, can anyone help me to optmized it further:

    Here is the code

    import os
import sys

MAX_EXP = 800

def make_two_power_set():
    return {1 << i for i in range(MAX_EXP+1)}

def count_matches(digits, powers):
    return sum(digits.count(str(power)) for power in powers)

if __name__ == '__main__':
    two_power_set = make_two_power_set()
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    t = int(sys.stdin.readline().rstrip())

    for t_itr in range(t):
        a = sys.stdin.readline().rstrip()

        result = count_matches(a, two_power_set)

        fptr.write(str(result) + '\n')

    fptr.close()