Google Interview Question for Software Engineers


Country: United States
Interview Type: Phone Interview




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gjgjg

- khkh December 02, 2019 | Flag Reply
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The idea is to map the 2d points to some sort of fixed reference and only compare where it make sense, instead of comparing each point with all the other point, which makes it a N2 algorithm. So here is my take on this. Divide the entire grid into a chunk of k x k grid squares.
e.g. 1st squre(0,0)(0,k)(k,0)(k,k)
2nd squre (0,k)(k,k)(0,2k)(k.2k)
Then iterate over each point and assign a parent grid
CEIL(i/k) * k, CEIL(j/k)*k

Once we have the squres assigned, iterate over the points which belongs only to adjacent 8 sqaures. Rest will lay out of bound for any 2d points inside this grid. Then like minimum spanning tree algorithm, keep chaining the points.

- Confused Aatma December 03, 2019 | Flag Reply
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Order : O(N)
However, if all the 2d points are packed with in these 9 squres, then a further division will reduce the number of iteration within the points.

- Confused Aatma December 03, 2019 | Flag


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