# sumHeights2 challenge

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.

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
LVL 7
###### Who is Participating?

Billing EngineerCommented:
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

Author Commented:
sumHeights  is not clear yet
0

Billing EngineerCommented:
ok, clarified on that other question
0

Author Commented:
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;
}
above passes all tests . Any improvements or alternate approaches?
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Billing EngineerCommented:
looks correct to me.
as mentioned in the other question, you should consider commenting the code
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Commented:
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;
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Author Commented:
: -(heights[j + 1] - heights[j])
i like this trick of multiplying with minus
0
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