July 2019 Lunchtime Discussion

70

Ab Div1 mai dekho kaise maje aate hai :stuck_out_tongue: :stuck_out_tongue:

71

Kya rank aaya ? @ankur314 @l_returns

72

Div1 is pretty ruthless… doesn’t care if you solve 1/2 questions even they were hard :frowning:

74

Mat pucho bro :cry::cry:

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75

Same situation bro…first qn was easy…but brain freeze hoggaya…last 10 minutes maai aaya solution but it was too late…agar first waala right kiya hota tumne aur maine top - 150 pakka tha :frowning:

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76

Next kickstart par focus krte hai ab!

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77

Next mai pel denge gaurantee.

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78

No editorials posted till now.

79

Ye bad news hai. Next time ratings nhi badhenge :joy::joy:(unrated)

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80

Muje pata tha soln :stuck_out_tongue:

81

It’s upto setter/admin now. Ask mentioning them.
I think editorialist did his work yesterday itself.

82

Kaha Aya. Kuch submit he nhi kiya :sweat_smile::joy:

83

I also knew but I wanted to solve all 3 so I got all anxious due to less practice of short contests…if there were 6-7 hours I would be very comfortable. :slight_smile:

84

if you are talking about the kickstart then can you share your idea on how to solve 2nd problem?

85

I solved all problems partially.
But,
Yup, I have solved 2 and a half problems after the contest.
For 2nd problem…I create a pile of discs and keep checking. Finally, I get the correct answer.
For the last one, I’ve reached the solution for it if abs() is not involved…will take more 1 hiur to crack it.
I did see you in the contest but forgot your rank :slight_smile:

86

HOW TO SOLVE THE QUESTION TWO LANES??

87

greedy works
make freq of contiguous obstacles in same lane <=2
Meaning if lanes are 1 1 1 2 2 2 2 1 2 1 2
Then make it 1 1 2 2 1 2 1 2 by keeping first and last obstacle of each type. Meaning we remove 2nd , 5th and 6th obstacle as they are extra.
Now it’s greedy. Think a bit more and you should get solution. Ping me if you don’t get it. I will give more hints.

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88

I was stuck on one particular thought yesterday … What if the obstacles are 1 2 1 and we are currently on 1st obstacle… Now let’s say that 2nd and 3rd obstacle didn’t have distance of d , so we will have to change our lane before we encounter obstacle 1 so that we are able to change at 2 … How does it deals with this situation?

89

I didn’t get your question but I will explain you the core logic.

   |p              |             :
        |      |        |        :

See now if he is a position p then he will switch lane at position \max(p+d,pos_3)
Now if \max(p+d,pos_3) \geq pos_4 (meaning if there is an obstacle in b/w then it’s not possible to move ahead.
Then it’s impossible.
If not then switch to the other lane and apply same logic again.
As soon as you switch lane apply - \max(p+d,pos_of_obstacle on the lane you want to switch to)
As I said just remove extra obstacles before applying above logic and keep
1 1 2 2 1 2 1 2 from 1 1 1 2 2 2 2 1 2 1 2
By removing obstacle at position 2,5,6.
Here is my AC solution applying above logic.
Also my code removes all obstacles which are after two obstacle in different lanes having difference of position=1 as we can’t go further.

1 Like
90

for(i=1;i<n;i++)
if(x[i-1].second+1==x[i].first){
k=x[i].first;
break;
}

is this piece really necessary … doesnt the second loop take care of this condition??