Your browser (Unknown 0) is no longer supported. Some parts of the website may not work correctly. Please update your browser.

UPCOMING CHALLENGES:

CURRENT CHALLENGES:

Molybdenum 2019

PAST CHALLENGES

Niobium 2019

Zirconium 2019

Yttrium 2019

Strontium 2019

Rubidium 2018

Arsenicum 2018

Krypton 2018

Bromum 2018

Future Mobility

Grand Challenge

Digital Gold

Selenium 2018

Germanium 2018

Gallium 2018

Zinc 2018

Cuprum 2018

Cutting Complexity

Nickel 2018

Cobaltum 2018

Ferrum 2018

Manganum 2017

Chromium 2017

Vanadium 2016

Titanium 2016

Scandium 2016

Calcium 2015

Kalium 2015

Argon 2015

Chlorum 2014

Sulphur 2014

Phosphorus 2014

Silicium 2014

Aluminium 2014

Magnesium 2014

Natrium 2014

Neon 2014

Fluorum 2014

Oxygenium 2014

Nitrogenium 2013

Carbo 2013

Boron 2013

Beryllium 2013

Lithium 2013

Helium 2013

Hydrogenium 2013

Omega 2013

Psi 2012

Chi 2012

Phi 2012

Upsilon 2012

Tau 2012

Sigma 2012

Rho 2012

Pi 2012

Omicron 2012

Xi 2012

Nu 2011

Mu 2011

Lambda 2011

Kappa 2011

Iota 2011

Theta 2011

Eta 2011

Zeta 2011

Epsilon 2011

Delta 2011

Gamma 2011

Beta 2010

Alpha 2010

ambitious

Find the longest valid slice of a sequence of brackets after performing at most K bracket rotations.

Programming language:

A bracket sequence is considered to be a valid bracket expression if any of the following conditions is true:

- it is empty;
- it has the form "
(U)" whereUis a valid bracket sequence;- it has the form "
VW" whereVandWare valid bracket sequences.

For example, the sequence "`(())()`" is a valid bracket expression, but "`((())(()`" is not.

You are given a sequence of brackets S and you are allowed to rotate some of them. Bracket rotation means picking a single bracket and changing it into its opposite form (i.e. an opening bracket can be changed into a closing bracket and vice versa). The goal is to find the longest slice (contiguous substring) of S that forms a valid bracket sequence using at most K bracket rotations.

Write a function:

class Solution { public int solution(String S, int K); }

that, given a string S consisting of N brackets and an integer K, returns the length of the maximum slice of S that can be transformed into a valid bracket sequence by performing at most K bracket rotations.

For example, given S = "`)()()(`" and K = 3, you can rotate the first and last brackets to get "`(()())`", which is a valid bracket sequence, so the function should return 6 (notice that you need to perform only two rotations in this instance, though).

Given S = "`)))(((`" and K = 2, you can rotate the second and fifth brackets to get "`)()()(`", which has a substring "`()()`" that is a valid bracket sequence, so the function should return 4.

Given S = "`)))(((`" and K = 0, you can't rotate any brackets, and since there is no valid bracket sequence with a positive length in string S, the function should return 0.

Write an ** efficient** algorithm for the following assumptions:

- string S contains only brackets: '
(' or ')';- N is an integer within the range [1..30,000];
- K is an integer within the range [0..N].

Copyright 2009–2019 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.