Best algorithm for delete duplicates in array of strings

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开发者_开发百科(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

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

You can often use a space-time tradeoff and invest more space to reduce time.

In this case you could use a hash table to determine the unique words.

add is O(n), so your CC calculation is wrong. Your algorithm is O(n^2).

Moreover, how would remove be implemented? It also looks like it would be O(n) - so the initial algorithm would be O(n^3).

Binary search will only work if the array you're searching is sorted. I guess that's not the case here, or you wouldn't be looping over your entire array in the inner loop of the original solution.

If the order of the final solution is irrelevant, you could break the array into smaller arrays based on length of the strings, and then remove duplicates from those arrays. Example:

// You have 
{"a", "ab", "b", "ab", "a", "c", "cd", "cd"}, 

// you break it into 
{"a", "b", "a", "c"} and {"ab", "ab", "cd", "cd"}, 

// remove duplicates from those arrays using the merge method that others have mentioned, 
// and then combine the arrays back together into 
{"a", "b", "c", "ab", "cd"}

This is the shortest algorithm that worked where arrNames and arrScores is parallel arrays and the highest score is taken.

I := 0;
J := 0;
//iCount being the length of the array

for I := 1 to iCount do
for J := I + 1 to iCount do

   if arrNames[I] = arrNames[J] then
   begin

     if arrScores[I] <= arrScores[J] then
     arrScores[I] := arrScores[J];

   arrScores[J] := arrScores[iCount];
   arrNames[J] := arrNames[iCount];

   arrScores[iCount] := 0;
   arrNames[iCount] := '';

   Dec(iCount);
   end;

def dedup(l):
    ht, et = [(None, None) for _ in range(len(l))], []
    for e in l:
        h, n = hash(e), h % len(ht)
        while True:
            if ht[n][0] is None:
                et.append(e)
                ht[n] = h, len(et) - 1
            if ht[n][0] == h and et[ht[n][1]] == e:
                break
            if (n := n + 1) == len(ht):
                n = 0
    return et