Is the single linked list already sorted, so identical elements are adjacent?

If its an sorted array that makes it easy.
But my question is incase of Unsorted array.
I know one method is liner search method. Where each indivual element is check once with all elements. But the complexity is O(n*n)

O-notation always covers the worst-case scenario. I don't think you can get an algorithm better than O(n*n) without sorting.

However, it is always possible to reduce the number of comparisons done. Things that work along the lines of a hash-table would probably work best.

AFAIK.

Ya that is what the solution i am looking for. Reduced number of comparisons and as you said can be done with code that goes along with Hash Table.
Actually I am looking for the Pseudo code atleast.

Basically, you use the hash to answer the "doesElementExist" test. If it doesn't, copy it to the unique list.

Or you could just insert the whole list into a balanced binary tree, then flatten it into a list when you're done. The inserts will be O(log2(n)) on average, either finding the item already present or finding where to insert it.

First of, i didn't know where to post this question as it isn't actually programming based but yeah. sorry.. Anyway, I was going through my binary file trying to figure ...

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