i think this prob cannot be solved by DP then...
Because anytime a new value coming into the non decreasing sub-array
obtained by the
DP equation can be disrupted(like in the above example).
I think a backtracking approach cud prove useful.
(Recursion)
anytime we get a new element we can do two
yes ..19 is the correct answer.
p(f1)
/ \
p1 p
nice...
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@ashish: cud u plzz explain a bit more...
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can you plzz tel me y -1 is interpreted as 11 ?
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@Yan Wang..can u tel me wat u meant by saying the following :
And we know all the different numbers in A[...]. we put them in a
rigorously increasing sequence B[1..p], where B[1] B[2] ...
B[p-1] B[p], and for every i from 1 to n, A[i] can be found in
B[1..p].
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@Yan Wang..can u tel me wat u meant by saying the following :
And we know all the different numbers in A[...]. we put them in a
rigorously increasing sequence B[1..p], where B[1] B[2] ...
B[p-1] B[p], and for every i from 1 to n, A[i] can be found in
B[1..p].
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@gene ..if u just give an example herethat will make things more
clear...
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