UKROBOT - Editorial

Okay, to be honest, that announcement is actually tripping me up. Sorry that it ended up so confusing, we probably should have just made it rows/columns instead of Cartesian coordinates.

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Hey man! Donā€™t worry. You are good.

Thatā€™s not an issue. Itā€™s okay.
Question quality is what matters which was quite good.

i really wonder how the tester tests these kind of problems

This checker actually seems like it would be simple. The constraints R, C \leq 20 are there for a reason - so the checker can brute-force all possible obstacle locations and compute the exact outcome for each one. If each outcome is different, then the solution is valid, otherwise, it may be possible that you canā€™t tell the difference, so the solution is wrong.

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so if generate random routes to nowhere each time ( ensuring they are distinct final ), still it would be accepted?

No, because the route has to be the same regardless of the obstacleā€™s position.

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my bad. Thanks for the reponse tho !!

which software you used to make this grid ?
please do tell me .

i think the question should be much more clear

The announcement and change in statement has to be wrong, right?

ok i am not getting it so let say if we have a grid (top left at 0,0) and obstacle at (1,1) and command is ā€œDRRRDā€ (D is down as (x,y)=>(x+1,y) and same others) so will the final pos will be (0,2) or not ??? if not then what will be the final position.

I canā€™t find why my approach is giving WA.
my code.
I think my code is mapping unique co-ordinate for each possible coordinate of obstacle.
From my approach let final coordinate of robot be (x,y).
Then, obstacleā€™s x co-ordinate will be= x and obstacleā€™s y co-ordinate will be = (y-r).
Help!

And one more thing :sweat_smile:, can solution for r=1 and c=20 be DRRRā€¦(R repeated 19 times) ??? here (0,0) is assumed at top left and r is row and c is col ,
D move make (x,y)=>(x+1,y)
and R make(x,y)=>(x,y+1)
because in this case all the final positions will be different for ex. let r=1 and c=3
the DRR
if obstacle at (1,1) final pos is (1,0)
if (1,2) -> final (1,1)
if (1,3)-> final (1,2)
as all are different so we have the answer and same for r=1 and c=20

or am i wrong somewhere ?

I was using the same approach. But still I didnā€™t understand it why it is not the valid solution

Wait, D should be (x, y) \to (x, y - 1), shouldnā€™t it?

I got the image on google lol. Just search for ā€œcartesian planeā€. It is the 6th image there. I used the native image editing software on my computer to draw straight red lines.

just for simplicity take D as x,y -> x+1,y
well you can take whatever you like i was comfortable in that so i wrote that but sry if i confused you :sweat_smile:

Can anyone tell me why this solution gets AC(for subtask 1) while this gets WA? shouldnā€™t we go up as C = 20? What am I missing?

The confusing part was that it would make D the same movement as R I think

Hereā€™s an example: your path is (0, 0) \to (0, 1) \to (1, 1) \to (1, 2) \to (2, 2), but the obstacle is in (1, 1). So youā€™d get to (0, 1), but the move to (1, 1) would be blocked. So you instead move to (0, 2) (ignore the previous move), then (1, 2).