• https://plot.ly/python/#fundamentals
• https://images.plot.ly/plotly-documentation/images/python_cheat_sheet.pdf

1. First solution first finds a boundary st.”01” and then extends the string to both sides to check if qualified.
2. Optimal Solution keeps counting how many contegious ‘0’ and ‘1’ currently and previously to achieve O(n) time and O(1) space.

It is a logic problem. My solution is:

1. If it contains 3 4 7, it is invalid.
2. If it contains only 0 1 8, it is valid but not different.
3. The rest cases are passed.

The quick and dirty solution is to use set/hashmap again. The other satisfied solution is using the sign of a number(+/-) to mark whether it shows up before, without extra space. Whoever came up with this really knows how to unitilize space…

Best solution is to use bitwise operation XOR: same ele will be cancel if xor twice.

BFS, one time AC

The key is to use no extra space, hence swap is most convenient. To optimize operation, only swap when encounter zeroes.

Easy Level.