Codility - Lesson 6 Sorting - 4. NumberOfDiscIntersections (better solution)

Source Link:
https://codility.com/programmers/lessons/6-sorting/number_of_disc_intersections/

Question:

We draw N discs on a plane. The discs are numbered from 0 to N − 1. A zero-indexed array A of N non-negative integers, specifying the radiuses of the discs, is given. The J-th disc is drawn with its center at (J, 0) and radius A[J].
We say that the J-th disc and K-th disc intersect if J ≠ K and the J-th and K-th discs have at least one common point (assuming that the discs contain their borders).
The figure below shows discs drawn for N = 6 and A as follows:
A[0] = 1 A[1] = 5 A[2] = 2 A[3] = 1 A[4] = 4
A[5] = 0

There are eleven (unordered) pairs of discs that intersect, namely:

• discs 1 and 4 intersect, and both intersect with all the other discs;
• disc 2 also intersects with discs 0 and 3.

Write a function:

class Solution { public int solution(int[] A); }

that, given an array A describing N discs as explained above, returns the number of (unordered) pairs of intersecting discs. The function should return −1 if the number of intersecting pairs exceeds 10,000,000.
Given array A shown above, the function should return 11, as explained above.
Assume that:

• N is an integer within the range [0..100,000];
• each element of array A is an integer within the range [0..2,147,483,647].

Complexity:

• expected worst-case time complexity is O(N*log(N));
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

My Solution:
Notes:
1. main idea:
// 1. store all the "lower points" and "upper points" of the discs
// 2. count the intersections (if one upper point > one lower point) 
2. use "long" for big numbers (be careful)
long[] lower = new long[A.length];
long[] upper = new long[A.length];
3. 
for(int i=0; i<A.length; i++){
lower[i] = i - (long)A[i]; // note: lower = center - A[i]
upper[i] = i + (long)A[i]; // note: upper = center + A[i]
}  
4. "sort" the "lower points" and "upper points"
Arrays.sort(upper);
Arrays.sort(lower);
5. scan the upper points
for(int i=0; i<A.length; i++){
// for the curent "j" (lower point)
while( j < A.length && upper[i] >= lower[j]){
intersection = intersection + j; // add j intersections
intersection = intersection - i; // minus "i" (avoid double count) 
j++;
}
}
6. for the overflow cases
if(intersection > 10_000_000)
return -1;
7. number of intersections 
return intersection;