lovro-sinda's blog

Task

By lovro-sinda, 10 years ago, In English

Can anyone help with this task.

Given a sequence of N different integers. Operations allowed members of the series are:

  • Move to the beginning of the sequence

  • Move to the end of the sequence

Print the smallest number of necessary operations to a default set of ascending sorted.

For example:

Input

4

3 4 2 1

Output

2

Thank you.

Tags dp, task
  • Vote: I like it
  • +5
  • Vote: I do not like it

Comments Comments (8)
| Write comment?
»
10 years ago, # |
  Vote: I like it +3 Vote: I do not like it

We move first 2 on the top

2 3 4 1

And in second move we move 1 on the top.

1 2 3 4

And answer is 2.

→ Reply
  • »
    »
    10 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    And why is it starting from "3 4 2 1" not "4 3 2 1"?

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Because sequence in the input is {3,4,2,1}.

      → Reply
      • »
        »
        »
        »
        10 years ago, # ^ |
        ← Rev. 2   Vote: I like it +5 Vote: I do not like it

        You've forgot to tell that in your example first 4 is the amount of remaining values, so your statement is bewildering.

        Usually to get answer you should provide as much clear and correct explanation as possible ;-)

        And by the way, what are limitations on input data? This may lead to different solutions...

        → Reply
»
10 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I think it was quite clear because text said ** different integers** . So with that assumption you could easily understand what is start sequence.

So I'll remember this suggestions for next time.

→ Reply
  • »
    »
    10 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    At first thought answer may be N - LIS where LIS is Longest Increasing Subsequence. LIS can be calculated with .

    → Reply
    • »
      »
      »
      10 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Doesn't work for case where sequence is {2,4,6,3,8,7}

      Answer by N- LIS is 2 , but correct answer is 3.

      → Reply
      • »
        »
        »
        »
        10 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        We have to maximise the set of elements that we don't move. At the end these will all be next to each other, so they have to be a substring of the sorted array, ordered correctly in the original array.

        This can be calculated in O(N) by finding, for each position in original array, the longest subsequence of original array ending at that position that is a substring of the sorted array. I think that overall it still must be O(NlogN) because you have to find the sorted array first.

        → Reply