geffen.nir
BAN USER
the hashing should work...first slope, than secondary hashing starting point of line on Y axis... if the same then up counter..
But it's still O(n^2) (finding all slopes is 2 of n, which makes it n(n-1) options).
and that does not include making the hash table. which could make it O(n^3), not to mention the O(n^2) in space...
no...there clearly has to be a simpler solution
you should sort it and then go over it from largest to smallest
if a[i] increase the gap then throw to 2nd array
else throw to first array.
O(nlogn+n)=O(nlogn)
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even if it's not binary, you still need the inorder.
- geffen.nir January 09, 2010else you don't know which part is to the left and which is to the right(binarically speaking).
you cannot recreate a tree from just it's postorder.