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;
}

Python without loop

def strangeCounter(t):
    # Write your code here
    r = math.ceil(math.log2(t/3+1)) # Calculate the number of current round
    sum_full = 3*(2**r-1) # Calculate the time when value reach to 1 at current round
    return sum_full - t + 1

long strangeCounter(long t) { long b = 1, tmp = 3, idx = 3, j; while (idx < t){ b += tmp; tmp *= 2; idx += tmp; } j = t - b + 1; return tmp - j + 1; }

def strangeCounter(t):! start=1 end=3 no = 3 ans=0 while(endend): break start = start+no no = no*2 end = end+no ans = 1+(end-t) return ans

return 6 * (2 ** math.floor(math.log2(1 + (t - 1) / 3))) - 2 - t