给定2-D平面中n个点的坐标，找出形成直角三角形的三元组的数量

Question

I can obviously do this by brute force, by selecting all the triplets one by one and checking if they form a right triangle. But what could be a more optimal way? I couldn't think of any.

edit: Since some of you pointed out another question seemingly similar to this one, I went through the answers and I still couldn't figure out how do I solve this one using those answers.

Answer 1

This will work in O(n^2Log(n)) .

Algo :

Create a 2D array Slope , where Slope[i,j] is slope vector between coordinate[i] and coordinate[j] .

Now, for coordinate[i] to be the joint (see note 2) of the right angled triangle , there should be two slopes in Slope[i,....] which are orthogonal to each other, ie their dot products are 0.

If Slope[i,j] = (a,b) (in vector form) , then you have to look for (b,-a) or (-b,a) in Slope[i,....] .

To make search optimized, keep the array Slope[i,...] sorted on a or b . Sorting will take O(nlogn) time.

Now, for any Slope[i,j] , do a binary search to look for (b,-a) and (-b,a) in array.

Binary Search for all n slopes will take O(nlogn) time.

So, total Time Complexity :

= Calculate slope between all points + Sort the slopes + BinarySearch 
= O(n*(n+nlogn+nlogn)) = O(n^2log(n)) 

Note :

1. Make sure to normalize the Slope Vector, ie :

   coordinate_a = (p,q) coordinate_b = (r,s) slope_vector(a,b) = (pr,qs)/distance(a,b)

2. Joint of Right Angled Triangle : Common coordinate between two orthogonal edges.