You can only view submissions for the problems you solved in the Gym.

The editorial will be available in 10 days as we're busy preparing for another contest.

for problem I you need to notice that the x coordinate and y coordinate of the optimal starting position are chosen independently cuz moving the robot start position horizontally never change how the robot can go out of the bounds vertically and vice versa/

Now to choose the optimal y pos for the robot you can notice that if you place the robot too high it can go out of the table from above and if you place it too low it can go out of the table from the bottom so the optimal is something in the middle so you can run ternary search and find the optimal y pos and you can do the same thing to find the optimal x pos too.

Note for problem I (Simple Robot):

it can be solved without ternary search and in a much faster way.

you can assume at the beginning that the optimal position is at point (0,0) (top left of the room). and then start executing the instructions sequentially.

let's talk about what could happen horizontally, if the next instruction you will skip is in < direction, you will consider changing the optimal x position by incrementing it by 1, but to do this you have to make sure of 2 things : that the current assumed optimal x position didn't reach the most right, and that the maximum right position you got to while executing the instructions is not the most right position, for the second condition you must keep track of the most right position you reached, and after incrementing the assumed optimal x position you have to increment the most right position you reached too. therefore this instruction that was to be skipped will not be skipped because you assumed starting at more right position.

but if the next instruction you will skip is either in:

1) > direction

2) < direction (and the current assumed optimal x position reached the most right, or the maximum right position you got to is the most right position)

then at this point the optimal x position doesn't need to be changed anymore and any instructions to be skipped starting from here (including the current instruction) will be skipped actually, so you count them and these will be the minimum horizontally.

for the vertical aspect you can think in the same way. (I solved it this way and it got accepted with time 109 ms.)

In lines 18, 19 you are doing N^2 iterations.