My simple solution of the order log n, it passed all test cases. My logic was that there was a pattern between the actual time and the time displayed on the stop watch. Am I missing something here? Many of the solutions here seem to take complex approaches.

rem = 3
while t > rem:
    t = t-rem
    rem *= 2

print(rem-t+1)

This is actually a math problem. Assume t is in the cycle with size 3*2^k, then k = floor(log2(t/3+1)). So, the problem is O(1).

A Java solution:

import java.util.Scanner;

public class StrangeCounter {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long t = in.nextLong(), n = 2;
        while (3 * (n - 1) < t) n = 2 * n;
        System.out.println((3 * (n - 1) - t + 1));
    }
}

Really implore fellow coders not to paste readymade solution in the board.

This is a real nice problem. If you can understand how the value is calculated and can find a pattern in there, you are already there.

here is the trick

long strangeCounter(long t) {
    long v=4;

    while(v<=t)
    {
        v=v*2+2;
    }
    return v-t;
}

just it :)