Compute the number of intersections in a sequence of discs.
- Difficulty Level: Medium
- Question URL: https://app.codility.com/programmers/lessons/6-sorting/number_of_disc_intersections/
- Time Complexity:
Question:
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:
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.
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].
Solution:
Solution to Codility's Number Of Disc Intersections problem which is from the Codility Lesson 6: Sorting and, is solved in Java 8 with 100% performance and correctness scores. The goal here is to compute the number of intersections in a sequence of discs. You can find the question of this NumberOfDiscIntersections problem in the Codility website.
package Codility.Lesson6;
import java.util.*;
public class NumberOfDiscIntersections {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] in2 = {8, 4, 5, 6,10,15,25 };
int result = solution(in2);
System.out.println(result);
}
public static int solution(int[] A) {
// write your code in Java SE 8
if (A.length < 3)
return 0;
Arrays.sort(A);
// made long array because each int element can be as high as
// Integer MAX_VALUE so when add them can overflow int
long[] aOrdered = new long[A.length];
int index = 0;
for (Integer i : A) {
aOrdered[index++] = i;
}
// start from the end (largest)
// if previous 2 elements have sum > current element, found a triangle
for (int i = aOrdered.length - 1; i >= 2; i--) {
if (aOrdered[i - 1] + aOrdered[i - 2] > aOrdered[i]) {
return 1;
}
}
return 0;
}
}