1766C - Hamiltonian Wall

Each column has at least 1 black, if all black through out, can return ‘YES’ straight
Otherwise, we need to visit each column once to ensure each black grid can be visited
The idea here is to find a starting point - the point here is only one black grid, so there is only one way to kickstart
Since the starting point can be in the middle, so we need to have 2 loops, one is to the left, another one to the right. We need to make sure all black grids at all columns can be visited and not repeated
When we move to a new column, we need to visit all the black in that column, and the first black must be adjacent to the previous one. If there are 2 black grids, and we only consider one is abit point, we need to come back to this column which is tough to program
So we just make sure we visit all black grids in current columns, and the last grid is going to be our new starting point → optimal structure(if we past each column, then we obtain the answer) & statelessness(to visit the next grid, we just need to know what is our current starting point)

Space - O(1), we are only using a few variables to deliver the answer
Time - O(n), we need to array to find the starting point, then visit each column once to ensure we can visit all black grids.