EZ Clap - Codeforces

#	User	Rating
1	jiangly	3845
2	tourist	3798
3	orzdevinwang	3706
4	jqdai0815	3682
5	ksun48	3589
6	Ormlis	3532
7	Benq	3468
8	Radewoosh	3463
9	ecnerwala	3451
10	Um_nik	3450

#	User	Contrib.
1	cry	165
2	Qingyu	160
3	-is-this-fft-	159
4	atcoder_official	157
5	Dominater069	156
6	adamant	154
7	djm03178	151
8	luogu_official	149
9	awoo	147
10	TheScrasse	145

1999/2000 ms

02:57:15 Unsuccessful hacking attempt
03:13:10 Unsuccessful hacking attempt
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03:13:53 Unsuccessful hacking attempt
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03:14:16 Unsuccessful hacking attempt
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13:36:27 Unsuccessful hacking attempt

Blessed by the $$$O(n^2 \log n)$$$ no-brainer! Try harder hacking next time :P

Comments (12)

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Aritro_

9 hours ago, # |

What is your approach that takes exactly $$$1999/2000$$$ ms?

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Leonardo_Paes

9 hours ago, # ^ |

← Rev. 2 →

I brute force the last painted element. That part is $$$O(n)$$$. Then, I get the subarrays from the left and from the right of it and sort them. The idea is that if I get at least one element on the left of our pivot and one from the right, I can guarantee that it is a valid last painted element. So, for $$$k\geq2$$$ I take the maximum on the left and on the right of our last painted element. Then, I sort all the other $$$n-3$$$ elements (without the last painted and the two maximums) and take the $$$k-2$$$ biggest ones. Sorting the remaining $$$n-3$$$ elements take $$$O(n \log n)$$$, so the overall complexity is $$$O(n^2 \log n)$$$. For $$$k = 1$$$, we just do a linear solution brute forcing the first painted item, noticing that the last one can only by the first or the last element of the array.

Here is the code!

DenimFlame12

6 hours ago, # ^ |

Same approach

Swayam78

8 hours ago, # |

Were you worried?

8 hours ago, # ^ |

Yes. During the contest it took 1984ms on pretests. I was considering implementing the $$$O(n)$$$ solution after getting D, but after 12 WAs I was a bit tilted and didn't have the patience to re-implement B.