Never mind. Figured it out, though in possibly different from the paper.
Principle of optimality to the rescue! :)
O(n^2)
DP equations:
LAP[i, j] = Longest AP in range [i,j] of the sequence
LAP[0, j] = max{LAP[0, k] (for all k in range [0, j-1]) extended with a[j],1
(the element a[j] itself)}
Result in LAP[0,n-1]
Make sure LAP maintains information about last element and common
difference.
(The simple cubic DP can be reduced to N^2 by observing that LAP[i,j] is
essentially the same as LAP[0, j] by principle of optimality as the result
needs to be calculated with start index 0).
@Sunny: Any bugs in the analysis?
--
DK
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http://www.divye.in
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