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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Strange Counter
  5. Discussions

Strange Counter

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992 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/Plsx1i8dqiI

    long strangeCounter(long t) {
    long last = 3, step = 3;
    while(t > last){
        step *= 2;
        last += step;
    }
    return last - t + 1;
}

  • maulanamlkib27
     + 0 comments

    ini solusi tanpa looping, dengan merumuskan masalah menjadi sebuah pola

        public static long strangeCounter(long t)
    {
        // 1, 4, 10, 22, 46 .... Un
        // 1 + (3* (2^0 - 1) => 1
        // 1 + (3* (2^1 - 1) => 4
        // 1 + (3* (2^2 - 1) => 10
        // 1 + (3* (2^3 - 1) => 22
        // 1 + (3* (2^4 - 1) => 46
        // 1 + (3* (2^5 - 1) => 94
        // ...
        // 1 + (3* (2^n - 1) => formula for the n-th term

        // how to get n from (2^n - 1), log2 (floor(time+2)/2)
        long n = (int)Math.Floor(Math.Log((t+2) / 3, 2));
        
        long topValue = long.Parse((3 * Math.Pow(2, n)).ToString());
        long topTime = topValue - 2;

        return topValue - (t - topTime);
    }

  • patrickgao2020
     + 0 comments

    Here is my Python code!

    def strangeCounter(t):
    time = 1
    value = 3
    while time + value <= t:
        time += value
        value = 2 * value
    return value - (t - time)

  • alekseybelousov
     + 0 comments

    The problem can be solved in O(1) If f(t) == 1, t = sum of geometric progression an = 3*2(n-1) . So we can find the next t where f(t) == 1

  • chauhansatish978
     + 0 comments
    long strangeCounter(long t) {
    long sum =3, counter = 3;
    while(sum < t){
        counter = counter*2;
        sum += counter;
    }
    
    return sum - t + 1;
}