Programming question

Question

Till now, there have been no COVID-19 cases identified in the country XYZ. Scientists and researchers there have conducted a research and discovered that, in a street with houses (imagine that the houses are only on one side of the street, the other side is empty), if residents in a house were affected with the novel Coronavirus, there is a high probability for the houses on either side of that house to be diagnosed with Coronavirus as well. This will result in an extremely high possibility for almost all the houses be affected by COVID-19.

Hearing this, the President of the country XYZ has decided to execute(kill) residents in some houses to save more people from that street.

You will be given an input string, containing the number of residents in adjacent houses for a specific street. Your task to find the maximum number of residents that can be saved from getting infected by COVID-19.

Input Format

Input string contains a set of integers seperated by a space



Constraints

Number of houses in a lane

Output Format

Print the maximum number of residents that can be saved from getting infected by COVID-19.

Sample Input 0

3 5 3 4 3 6

Sample Output 0

15

Answer 1

Hey Ashley you need to write a method that takes in a String. The string is your input.

Your constraint where it's less than 1000 and may be 0 means that you need to evaluate corner cases. Like what will you do if there are 0 houses.

Also, remember, sometimes, you may have to skip multiple houses to save the most number of people.

ex: 10 1 2 11

Here output should be 21.

Answer 2

Sounds like you add up the even # houses ---> even parity

Then you add up the odd # houses --> odd parity

class ArrayParity

{

public static void main( String args[])

{

if (args.length>2)

{

int N = args.length;

int [] A = new int[N];

for (int iLoop=0; iLoop<N; iLoop++)

{

A[iLoop] = Integer.parseInt(args[iLoop]);

//System.out.println(A[iLoop]);

}

int even_sum=0;

int odd_sum=0;

for (int iLoop=0; iLoop<N; iLoop+=2)

{

even_sum += A[iLoop];

}

for (int iLoop=1; iLoop<N; iLoop+=2)

{

odd_sum += A[iLoop];

}

System.out.println(" total of evens "+ even_sum);

System.out.println(" total of odds " + odd_sum);

}

}

}