Codility--- NumberOfDiscIntersections

Posted 2020-09-15 Hugh

Task description 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 show code in pop-up 12345678910111213141516171819202122232425262728293031323334353637383940414243444546 // 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; } }

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