Question

Today at school the teacher asked us to implement a duplicate-deletion algorithm. It’s not that difficult, and everyone came up with the following solution (pseudocode):

for i from 1 to n - 1
    for j from i + 1 to n
        if v[i] == v[j] then remove(v, v[j])    // remove(from, what)
    next j
next i

The computational complexity for this algo is n(n-1)/2. (We’re in high school, and we haven’t talked about big-O, but it seems to be O(n^2)). This solution appears ugly and, of course, slow, so I tried to code something faster:

procedure binarySearch(vector, element, *position)
    // this procedure searches for element in vector, returning
    // true if found, false otherwise. *position will contain the
    // element's place (where it is or where it should be)
end procedure

----

// same type as v
vS = new array[n]

for i from 1 to n - 1
    if binarySearch(vS, v[i], &p) = true then
        remove(v, v[i])
    else
        add(vS, v[i], p)      // adds v[i] in position p of array vS
    end if
next i

This way vS will contain all the elements we’ve already passed. If element v[i] is in this array, then it is a duplicate and is removed. The computational complexity for the binary search is log(n) and for the main loop (second snippet) is n. Therefore the whole CC is n*log(n) if I’m not mistaken.

Then I had another idea about using a binary tree, but I can’t put it down.
Basically my questions are:

• Is my CC calculation right? (and, if not, why?)
• Is there a faster method for this?

Thanks

Answer 1

The easiest solution will be to simply sort the array (takes O(n log n) with standard implementation if you may use them. otherwise consider making an easy randomized quicksort (code is even on wikipedia)).

Afterwards scan it for one additional time.
During that scan simple eliminate consecutive identical elements.

If you want to do it in O(n), you can also use a HashSet with elements you have already seen.
Just iterate once over your array, for each element check if it is in your HashSet.

If it isn’t in there, add it.
If it is in there, remove it from the array.

Note, that this will take some additional memory and the hashing will have a constant factor that contributes to your runtime. Althought the time complexity is better, the practical runtime will only be onyl be faster once you exceed a certain array size