## CCC '15 S3 - Gates

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Points: 10 (partial)
Time limit: 1.0s
Python 2 2.5s
Python 3 2.5s
Memory limit: 256M

Author:
Problem type
##### Canadian Computing Competition: 2015 Stage 1, Senior #3

For your birthday, you were given an airport.

The airport has gates, numbered from to . planes arrive at the airport, one after another. You are to assign the plane to permanently dock at any gate (), at which no previous plane has docked. As soon as a plane cannot dock at any gate, the airport is shut down and no future planes are allowed to arrive.

In order to keep the person who gave you the airport happy, you would like to maximize the number of planes starting from the beginning that can all dock at different gates.

#### Input Specification

The first line of input contains (), the number of gates at the airport.

The second line of input contains (), the number of planes which will land.

The next lines contain one integer , (), such that the plane must dock at some gate from to , inclusive.

Note that for at least of the marks for this question, and .

#### Output Specification

Output the maximum number of planes that can land starting from the beginning.

#### Sample Input 1

4
3
4
1
1

#### Output for Sample Input 1

2

#### Explanation of Output for Sample Input 1

The first plane can go anywhere, but it is best to not put it into Gate . Notice that planes and both want to dock into Gate , so plane is unable to dock.

#### Sample Input 2

4
6
2
2
3
3
4
4

#### Output for Sample Input 2

3

#### Explanation of Output for Sample Input 2

The first two planes will dock in gates and (in any order). The third plane must dock at Gate . Thus, the fourth plane cannot dock anywhere, and the airport is closed, even though plane would have been able to dock.

• commented on Sept. 19, 2021, 11:34 p.m.

sick birthday present

• commented on Feb. 10, 2021, 10:14 a.m.

hi can someone give me a hint to how to implement an equivalent of the std::bitset's _find_next() function in python?

• commented on June 30, 2021, 3:01 p.m.

You should implement a solution with the correct complexity instead. Bitset is not the best way to approach this problem.

• commented on Jan. 19, 2021, 10:33 p.m.

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• commented on Jan. 20, 2021, 5:28 p.m.

• commented on Jan. 20, 2021, 4:26 p.m.

A plane 'i' can be assigned to any airport from g1 to gi. It seems like your code above just automatically assigns it to gi. In the case of:

4 4 4 4 4 4

all planes can dock, but your code would output 1. You also have to consider that the airport shuts down after a plane can no longer dock.

In terms of time complexity, the code above is O(n^2), which doesn't pass the limit.

• commented on Dec. 27, 2020, 4:58 p.m.

For Batch #11 Case #3, I got a TLE; however, on the official CCC grader, my submitted code gets full marks since the time limit is 5 seconds. So are the problems here supposed to be mirror's of the one's found on other websites? Or could there be minor differences like time limit?

• commented on Dec. 27, 2020, 5:35 p.m. edit 2

In the second set of cases they augmented the test data to eliminate weaker solutions (That can still pass on the CCC). You have to find a more efficient solution to pass the second set of cases, try looking at the data structure set.

• commented on Dec. 22, 2020, 7:27 p.m.

What kind of person gets an airport as their birthday...

• commented on Dec. 10, 2020, 4:31 p.m. edit 3

I think my solution is broken, but it still passes. Using this test case

5
5
1
2
3
5
5

my solution outputs 4, even though it should be 5. Did I misunderstand something, or has my solution miraculously passed despite being wrong?

Edit: just solved the problem the right way, the solution I am referring to above can be found at https://dmoj.ca/submission/3196239

• commented on May 20, 2020, 12:31 p.m.

Can anyone tell me why std::lower_bound TLEs, but set::lower_bound ACs? I thought they both ran in O(logN)

• commented on May 20, 2020, 5:23 p.m.

set::lower_bound is a specialized lower_bound method for std::set that runs in time. std::lower_bound only has this guarantee for random access iterators, which std::set does not have. For non random access iterators such as with std::set, it takes time on average (based on http://www.cplusplus.com/reference/algorithm/lower_bound/).

• commented on July 18, 2018, 3:34 p.m.

La-da-da-da-dah You know I'm mobbin' with the T.L.E.

• commented on June 29, 2021, 10:05 a.m.

make this a real remix of the song

• commented on Jan. 11, 2017, 9:21 p.m.

Even the O(n ^ 2) Algorithm worked as long as you code it in C

• commented on Jan. 13, 2017, 7:08 p.m.

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• commented on Jan. 16, 2017, 11:17 a.m.

I still passed lol.

• commented on Jan. 4, 2017, 7:52 p.m.

• commented on Aug. 9, 2016, 12:04 p.m.

How come my 15/30 submission did not get TLE.

• commented on Aug. 11, 2016, 12:34 p.m. edited

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• commented on Aug. 6, 2016, 9:15 a.m.

I think Java is missing in the allowed languages.

• commented on Aug. 6, 2016, 3:05 p.m.

Thanks for letting us know, it's been added now.

• commented on May 17, 2016, 10:26 p.m.

Why is there no judge for this problem?

• commented on Aug. 3, 2015, 4:44 a.m.

Is it advised not to use ArrayList as it is slower than a regular array to optimize the speed of the program? Is the difference between an array and an ArrayList not large enough to slow down the program so drastically that the time limit would not be exceeded? If not, then it must be the algorithm that is at fault then, correct? And if so, any tips on how to optimize my algorithm?

• commented on Aug. 3, 2015, 11:19 a.m.

Your algorithm is wrong, one possible approach is to use binary search and a set to reduce to run time complexity to

Not sure what the set equivalent is in java though.

• commented on May 18, 2015, 9:45 a.m.

TIL that .

• commented on Feb. 22, 2015, 1:44 p.m. edited

The original CEMC test data is now a 50% subtask (5 points). New cases, worth the other 50%, are added to award to the truly correct solutions.

All submissions have been rejudged.

Note: For of the CEMC cases, and . Those cases are worth of the total points.