Codility--- NumberOfDiscIntersections

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.

Solution

Programming language used: Java

Total time used: 4 minutes

Code: 15:38:12 UTC, java, final, score:  100

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// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
        int l = A.length;
        int[] arrayIn = new int[l];
        int[] arrayOut = new int[l];
        int inNumContext = 0;
        int result = 0;  

        for(int i = 0; i < l; i ++) {
            int in = (i - A[i]) < 0 ? 0 : (i - A[i]);
            // take care the (A[i] + i) exceeds the value of MAX int
            // which will become minus int
            int out = (A[i] + i > l - 1 || A[i] + i < 0) ? (l - 1) : (A[i] + i);
            arrayIn[in] ++;
            arrayOut[out] ++;
        }  

        for(int i = 0; i < l; i ++) {
            if(arrayIn[i] != 0) {
                // previous circles times new coming circles
                result += inNumContext * arrayIn[i];
                // new coming circles group with each other
                result += arrayIn[i] * (arrayIn[i] - 1) / 2;  

                if (result > 10000000) {
                    return -1;
                }  

                // add coming circles to inNumContext
                inNumContext += arrayIn[i];
            }
            // minus leaving circles from inNumContext
            inNumContext -= arrayOut[i];
        }   

        return result;
    }
}