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:

Pi Code Challenge

PAST CHALLENGES

Year of the Rabbit

Carol of the Code

Game of Codes

National Coding Week 2022

Jurassic Code

Fury Road

Bug Wars: The Last Hope

Muad'Dib's

Year of the Tiger

Pair a Coder

Code Alone

Gamer's

Spooktober

National Coding Week

The Coder of Rivia

Fast & Curious

The Fellowship of the Code

May the 4th

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

A matrix of binary values is given. We can flip the values in selected columns. What is the maximum number of rows that we can obtain that contain all the same values?

Matrix A, consisting of N rows and M columns, is given, with each cell containing the value 0 or 1. Rows are numbered from 0 to N−1 (from top to bottom). Columns are numbered from 0 to M−1 (from left to right). The values inside the matrix can be changed: you can select as many columns as you want, and in the selected column(s), every value will be flipped (from 0 to 1, or from 1 to 0).

The goal is to obtain the maximum number of rows whose contents are all the same value (that is, we count rows with all 0s and rows with all 1s).

Write a function:

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

that, given matrix A, returns the maximum number of rows containing all the same values that can be obtained after flipping the selected columns.

**Examples:**

1. Given matrix A with N = 3 rows and M = 4 columns:

the function should return 2. After flipping the values in column 1, the two last rows contain all equal values. Row 1 contains all 0s and row 2 contains all 1s.

2. Given matrix A with N = 4 rows and M = 4 columns:

the function should return 4. After flipping the values in two of the columns (columns 0 and 2), all the rows have the same value. Rows number 0 and 2 contain all 1s, and rows number 1 and 3 contain all 0s.

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

- N and M are integers within the range [1..100,000];
- N * M is not greater than 100,000.

Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.