Google Interview Question for SDE-2s


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2
of 2 vote

observation:
after 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.

- plarion October 27, 2019 | Flag Reply
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0
of 0 votes

Also, we can find out what point groups can not mutually create a circle. How? Simple, any 3 points on a plane can be made into a circle, unless all 3 of them form a line.

It is O(n^2) to check all points to find out all lines, at most.
So, we would have point groups which can not be used as circle.
That automatically should reduce a lot of points comparison.

They, might be possibly more optimisations, but am thinking still. Hence a comment, not an answer.

- NoOne October 30, 2019 | Flag
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0
of 0 vote

I am tempted to say it is not possible to avoid O(n^3) in general. Consider a case where you have N points but only 3 of them lie on a circle (i.e. getCircle returns non-null). Now you cannot find the circle unless you test all triplets.

- Anonymous October 27, 2019 | Flag Reply
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0
of 0 vote

Look into Welzl's algorithm it's a O(log n) implementation for finding the center of a circle using Randomization.

- Anonymous October 28, 2019 | Flag Reply
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0
of 0 vote

What would be the recursive algorithm for this?

- Ubiquetous November 02, 2019 | Flag Reply
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0
of 0 vote

I think getCircle already returns center point and radius right. Say it returns center point as (p,q), and radius = r

After finding the triplet, for each other point in the list
square-root of ((x-p)^2 + (y-q)^2) == radius will give the points in that circle perimeter.
Buffer all these points in a list.

Repeat above for other triplet combinations, but skip the combination where all three points are already part of a circle identified.

- Chaitanya Srikakolapu November 27, 2019 | Flag Reply
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0
of 0 votes

Won't this be O(N4). O(N3) for each combination and iterating over all the point once again to calculate the distance.

- Confused Aatma November 27, 2019 | Flag
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0
of 0 vote

I can think an optimization where we create a fully connected graph, where the edge weight represent distance between 2 points sqrt((a2-a1)2 + (b2-b1)2). Store these edges in a sorted order list. This would take O(n2). Now when we choose triplets and get a circle, we start iterating over nearest neighbor from each of 3 points, till a stage where the distance is less than diameter. This way when we ignore all the points which will definitely be out of circle.

- Confused Aatma November 27, 2019 | Flag Reply


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