the problem seems to have many cases a>0, a=0,a<0, (when a=0, b>0, b<0). However, they can be combined into just 2 cases: a>0 or a<0

1.a>0, two ends in original array are bigger than center if you learned middle school math before.

2.a<0, center is bigger than two ends.

so use two pointers i, j and do a merge-sort like process. depending on sign of a, you may want to start from the beginning or end of the transformed array. For a==0 case, it does not matter what b's sign is.

The function is monotonically increasing or decreasing. you can start with either beginning or end.

```
public class Solution {
public int[] sortTransformedArray(int[] nums, int a, int b, int c) {
int n = nums.length;
int[] sorted = new int[n];
int i = 0, j = n - 1;
int index = a >= 0 ? n - 1 : 0;
while (i <= j) {
if (a >= 0) {
sorted[index--] = quad(nums[i], a, b, c) >= quad(nums[j], a, b, c) ? quad(nums[i++], a, b, c) : quad(nums[j--], a, b, c);
} else {
sorted[index++] = quad(nums[i], a, b, c) >= quad(nums[j], a, b, c) ? quad(nums[j--], a, b, c) : quad(nums[i++], a, b, c);
}
}
return sorted;
}
private int quad(int x, int a, int b, int c) {
return a * x * x + b * x + c;
}
}
```