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observation:
- plarion October 27, 2019after you calculate a circle - find all points on the circle (o(n))
since a triplet form only one circle - so all combination of point on that circle form the same circle.
so there is no need to re-calculate a circle with them.
storing them in a hash and skipping them in the future will reduce run-time.
you will still go over all triplets (o(n^3)) but you will call the method getCircle(...) much fewer times.