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

UPCOMING CHALLENGES:

Ruthenium 2019

CURRENT CHALLENGES:

Technetium 2019

PAST CHALLENGES

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

ambitious

While removing edges from a mesh grid, find the moment when there ceases to be a connection between opposite corners.

Programming language:

There is an N **×** N square mesh-shaped grid of wires, as shown in a figure below. Nodes of the grid are at points (X, Y), where X and Y are integers from 0 to N−1. An electric current flows through the grid, between the nodes at (0, 0) and (N−1, N−1).

Initially, all the wires conduct the current, but the wires burn out at a rate of one per second. The burnouts are described by three arrays of integers, A, B and C, each of size M. For each moment T (0 ≤ T < M), in the T-th second the wire between nodes (A[T], B[T]) and:

- (A[T], B[T] + 1), if C[T] = 0 or
- (A[T] + 1, B[T]), if C[T] = 1

burns out. You can assume that the arrays describe existing wires, and that no wire burns out more than once. Your task is to determine when the current stops flowing between the nodes at (0,0) and (N−1,N−1).

Write a function:

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

that, given integer N and arrays A, B and C, returns the number of seconds after which the current stops flowing between the nodes at (0, 0) and (N−1, N−1). If the current keeps flowing even after all M wires burn out, the function should return −1.

For example, given N = 4, M = 9 and the following arrays:

your function should return 8, because just after the eighth wire burns out, there is no connection between the nodes at (0, 0) and (N−1, N−1). This situation is shown in the following figure:

Given N = 4, M = 1 and the following arrays:

your function should return −1, because burning out a single wire cannot break the connection between the nodes at (0, 0) and (N−1, N−1).

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

- N is an integer within the range [1..400];
- M is an integer within the range [0..2*N*(N−1)];
- each element of arrays A, B is an integer within the range [0..N−1];
- each element of array C is an integer that can have one of the following values: 0, 1.

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