1.My idea is pretty simple, since we encounter 1, we just pop out 1 + 'x', if we encounter 0
we just pop 0 itself, if the remaining bits == [0], then it means we get the correct answer.
Note: the pop(0) takes O(n) time, hence this algothm takes O(n^2) time and O(1) space
2.A natural idea to think is to represent bits using a queue bits_queue, since bits_queue.popleft() takes O(1) time.
Note: this algorithm takes O(n) time and O(n) space
3.We just create a flag length to illustrate how many digits we have already passed, if we encounter
1, then length += 2, else length += 1. finally we compare length = len(bits)-1 .
Note: this algothm takes O(n) time and O(1) space