Solved

sumHeights2  challenge

Posted on 2016-08-30
7
146 Views
Last Modified: 2016-09-08
Hi,


i am working one below challenge
http://codingbat.com/prob/p157900


i was not clear on below description




(A variation on the sumHeights problem.) We have an array of heights, representing the altitude along a walking trail. Given start/end indexes into the array, return the sum of the changes for a walk beginning at the start index and ending at the end index, however increases in height count double. For example, with the heights {5, 3, 6, 7, 2} and start=2, end=4 yields a sum of 1*2 + 5 = 7. The start end end index will both be valid indexes into the array with start <= end.

how below results are 7,2, 15 each? please advise

sumHeights2([5, 3, 6, 7, 2], 2, 4) → 7
sumHeights2([5, 3, 6, 7, 2], 0, 1) → 2
sumHeights2([5, 3, 6, 7, 2], 0, 4) → 15
0
Comment
Question by:gudii9
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 3
7 Comments
 
LVL 143

Accepted Solution

by:
Guy Hengel [angelIII / a3] earned 500 total points
ID: 41777525
indeed, if you have the sumHeights done, this one becomes very similar.
the only difference is that if the value goes "up", you have to double it, hence the simple function "absolute" will not do the job, you need to take the difference and check it's value before going on with the eventual " *2  " before adding the result to the total
0
 
LVL 7

Author Comment

by:gudii9
ID: 41780970
sumHeights  is not clear yet
0
 
LVL 143

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 41781143
ok, clarified on that other question
0
What Is Transaction Monitoring and who needs it?

Synthetic Transaction Monitoring that you need for the day to day, which ensures your business website keeps running optimally, and that there is no downtime to impact your customer experience.

 
LVL 7

Author Comment

by:gudii9
ID: 41788975
public int sumHeights2(int[] heights, int start, int end) {
 int diff=0;
  int sum=0;
  int absDiff=0;
  for(int i=start; i<end; i++){
    
    diff=heights[i+1]-heights[i];
    if(diff<0){
    absDiff=Math.abs(diff);
    }else{
      absDiff=2*diff;
    }
    sum=sum+absDiff;
  }
  return sum;
}

Open in new window

above passes all tests . Any improvements or alternate approaches?
0
 
LVL 143

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 41789071
looks correct to me.
as mentioned in the other question, you should consider commenting the code
0
 
LVL 32

Expert Comment

by:awking00
ID: 41790239
Alternative method without absolute -
int diff=0;
int sum = 0;
for(int j=start; j<end; j++) {
      diff = (heights[j + 1] > heights[j]) ? 2*(heights[j + 1] - heights[j]) : -(heights[j + 1] - heights[j]);
      sum = sum + diff;
}
return sum;
0
 
LVL 7

Author Comment

by:gudii9
ID: 41790490
: -(heights[j + 1] - heights[j])

Open in new window

i like this trick of multiplying with minus
0

Featured Post

Online Training Solution

Drastically shorten your training time with WalkMe's advanced online training solution that Guides your trainees to action. Forget about retraining and skyrocket knowledge retention rates.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

In this post we will learn how to connect and configure Android Device (Smartphone etc.) with Android Studio. After that we will run a simple Hello World Program.
This article will inform Clients about common and important expectations from the freelancers (Experts) who are looking at your Gig.
In this fourth video of the Xpdf series, we discuss and demonstrate the PDFinfo utility, which retrieves the contents of a PDF's Info Dictionary, as well as some other information, including the page count. We show how to isolate the page count in a…
Starting up a Project

707 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question