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[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:
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].
方法一：Thiago Papageorgiou 大大提供的做法
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 ```
完整練習題 source code 請參閱：github
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