### Number Of Disc Intersections

January 28, 2020

We draw N discs on a plane. The discs are numbered from 0 to N − 1. An 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.

Write an efficient algorithm for the following assumptions:

• 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].

Hints:

- convert 2D discs into 1D line segments on X axis into
**touples**with Left-x and Right-x ([x0, x1]) - sort the touples by Left-x (x0)
- add counters if next segment's Left-x <= previous segment's Right-x

```
function solution(A) {
// write your code in JavaScript (Node.js 8.9.4)
//console.log(A)
//[1, 5, 2, 1, 4, 0]
// convert 2D discs into 1D line segments on X axis
let tupples = []
let i
let j
for(i=0; i<A.length; i++){
tupples.push( [ i-A[i], i+A[i] ] )
}
//console.log("source: ",tupples)
//[ [ -1, 1 ], [ -4, 6 ], [ 0, 4 ], [ 2, 4 ], [ 0, 8 ], [ 5, 5 ] ]
tupples.sort((a,b)=>a[0]-b[0])
//console.log("sorted: ", tupples)
//[ [ -4, 6 ], [ -1, 1 ], [ 0, 4 ], [ 0, 8 ], [ 2, 4 ], [ 5, 5 ] ]
let count = 0
for(i=0; i<A.length; i++){
for(j=i+1; j<A.length; j++){
if(tupples[j][0] <= tupples[i][1]){
// next j segment start point is inside i segment
count ++
if(count>10000000) return -1
}else{
break
// next j segments are not inside i segment
}
}
}
return count
}
```