NumberOfDiscIntersections.md

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:

function solution(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.

function sortAsc(a, b) {
 return (a - b)
}

function solution(A) {
  var counter = 0, j= 0;
  var lower = [];
  var upper = []

  for(var i=0; i < A.length; i++) {
   lower[i] =i-A[i];
   upper[i] =i+A[i];
  } 

  lower.sort(sortAsc)
  upper.sort(sortAsc)

  for(var i= 0; i<A.length; i++) { 
   while(j < A.length && upper[i] >= lower[j]){
    counter += j-i;
    j++; 
   } 
  	if(counter > 10000000) return -1;
  } 

  return counter;
}