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Codility--- NumberOfDiscIntersections

2024-09-18 14:37:18   216人阅读

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

技术分享

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;    }}



https://codility.com/demo/results/training5N9W8K-3M3/

Codility--- NumberOfDiscIntersections


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