##
**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; }

public int sumHeights2(int[] heights, int start, int end) {

ReplyDeleteint 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;

}

public int sumHeights2(int[] heights, int start, int end) {

ReplyDeleteint 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

public int bigHeights(int[] heights, int start, int end) {

ReplyDeleteint 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;

}