## 2017年2月21日 星期二

### 【Python】Codility in Python : Lesson 6 - Sorting【NumberOfDiscIntersections】

Sorting 第四題：【NumberOfDiscIntersections】
Compute the number of intersections in a sequence of discs.
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 = 1
A = 5
A = 2
A = 1
A = 4
A = 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:

def 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.

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].

Correctness：100%、Performance：100%
```python upper = sorted([k + v for k, v in enumerate(A)]) lower = sorted([k - v for k, v in enumerate(A)]) j = 0 counter = 0 for i, v in enumerate(upper): while j < len(upper) and v >= lower[j]: counter += j-i j += 1 if counter > 10**7 : return -1 return counter ```