Faster algorithm

[qweasdzxcqweasd]

I need faster algorithm. Problem - to leave only the last occurrences of each element in the sequence; the remaining elements must be removed from the sequence. Thank you.

Given: 1 2 3 3 2 1 4 1 2 0
Answer:3 4 1 2 0
My algorithm:

k = input().split()  
    s=[] 
    for i in k:
        if i not in s:
          s.append(i)
        else:
          s.remove(i)  
          s.append(i)

[deanhystad]

Stop thinking about python for a minute. This is an easy problem.

Write down a list on a piece of paper. What is the last number in the answer?

Output:
1 2 3 3 2 1 4 1 2 0
                  ^
Answer = 0

What number do you check next. Is it added to the answer?

Output:
1 2 3 3 2 1 4 1 2 0
                ^
Answer = 2 0

This is repeated for 1, 4, 1, 2, 3, 3, 2, 1
What happens when we encounter a number that is already in the answer?
output]
1 2 3 3 2 1 4 1 2 0
^
Answer = 4 1 2 0[/output]
1 is already in Answer, so it does not get added to the answer list.

That is the algorithm. Now all you need to do is convert it to code. I wonder if python has a way to get items from a list starting at the end? I wonder if python has a way to insert items at the front of a list?



The next n

[jefsummers]

Another thought - from your description you want just the unique values. A set is a collection with unique values. Convert the list to a set, then back again. Voila. Bet that will be really fast.