Problem:

(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

Solution:

public int sumHeights2(int[] heights, int start, int end) {
int tmp = 0;
for (int i = start; i <= end-1; i++) {
if (heights[i] < heights[i+1]) {
tmp += 2 * (Math.abs(heights[i] - heights[i+1]));
}
else
tmp += Math.abs(heights[i] - heights[i+1]);
}
return tmp;
}

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

2. public int sumHeights2(int[] heights, int start, int end) {
int sumHeight = 0;
for(int i = start; i < end; i++)
{
sumHeight += (heights[i] < heights[i+1]) ? 2*(heights[i+1] - heights[i]) : heights[i] - heights[i+1];
}
return sumHeight;
}
Ternary is useful

3. public int bigHeights(int[] heights, int start, int end) {
int counter = 0;
for(int i = start; i < end; i++) {
if(heights[i] < heights[i+1]) {
if(heights[i+1] - heights[i] >= 5) {
counter++;
}
}
if(heights[i] > heights[i+1]) {
if(heights[i] - heights[i+1] >= 5) {
counter++;
}
}
}
return counter;
}