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:

May the 4th Challenge

PAST CHALLENGES

The Great Code Off 2021

The Doge 2021

The Matrix 2021

The OLX Group challenge

Silver 2020

Palladium 2020

Rhodium 2019

Ruthenium 2019

Technetium 2019

Molybdenum 2019

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

Find the maximum number of ropes that can be attached in order, without breaking any of the ropes.

In a room there are N ropes and N weights. Each rope is connected to exactly one weight (at just one end), and each rope has a particular durability − the maximum weight that it can suspend.

There is also a hook, attached to the ceiling. The ropes can be attached to the hook by tying the end without the weight. The ropes can also be attached to other weights; that is, the ropes and weights can be attached to one another in a chain. A rope will break if the sum of weights connected to it, directly or indirectly, is greater than its durability.

We know the order in which we want to attach N ropes. More precisely, we know the parameters of the rope (durability and weight) and the position of each attachment. Durabilities, weights and positions are given in three arrays A, B, C of lengths N. For each I (0 ≤ I < N):

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,
- C[I] (such that C[I] < I) is the position to which we attach the I-th rope; if C[I] equals −1 we attach to the hook, otherwise we attach to the weight connected to the C[I]-th rope.

The goal is to find the maximum number of ropes that can be attached in the specified order without breaking any of the ropes.

Write a function:

class Solution { public int solution(int[] A, int[] B, int[] C); }

that, given three arrays A, B, C of N integers, returns the maximum number of ropes that can be attached in a given order.

For example, given the following arrays:

the function should return 3, as if we attach a fourth rope then one rope will break, because the sum of weights is greater than its durability (2 + 3 + 1 = 6 and 6 > 5).

Given the following arrays:

the function should return 2, as if we attach a third rope then one rope will break, because the sum of weights is greater than its durability (2 + 2 + 1 = 5 and 5 > 4).

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

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.