Hi,
Consider the sequence 1,2,3,4,5,6...,n. There are n^2 blocks, so the
question may not be correct.
Regards
Aditya Shankar
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I might be wrong here but, why can't you just sort the block A[i..j]
it will take O((j-i)log(j-i)) (there are many O(n logn) sorting
algorthms) steps and then just look if they are in sequence trivially
another j-i steps.
On Dec 3, 1:33 am, Vinoth Kumar wrote:
> I kinda need the worst case also t
