How to find minimal-length subsequence that contains all element of a sequence

Question

Given a sequence such as S = {1,8,2,1,4,1,2,9,1,8,4}, I need to find the minimal-length subsequence that contains all element of S (no duplicates, order does n

Answer 1

I've got a O(N*M) algorithm where N is the length of S, and M is the number of elements (it tend to works better for small values of M, i.e : if there are very few duplicates, it may be a bad algorithm with quadratic cost) Edit : It seems that in fact, it's much closer to O(N) in practise. You get O(N*M) only in worst case scenarios

Start by going through the sequence and record all the elements of S. Let's call this set E.

We're going to work with a dynamic subsequence of S. Create an empty map M where M associates to each element the number of times it is present in the subsequence.

For example, if subSequence = {1,8,2,1,4}, and E = {1, 2, 4, 8, 9}

• M[9]==0
• M[2]==M[4]==M[8]==1
• M[1]==2

You'll need two index, that will each point to an element of S. One of them will be called L because he's at the left of the subsequence formed by those two indexes. The other one will be called R as it's the index of the right part of the subsequence.

Begin by initializing L=0,R=0 and M[S[0]]++

The algorithm is :

While(M does not contain all the elements of E)
{
    if(R is the end of S)
      break
  R++
  M[S[R]]++ 
}
While(M contains all the elements of E)
{
  if(the subsequence S[L->R] is the shortest one seen so far)
    Record it
  M[S[L]]--
  L++
}

To check if M contains all the elements of E, you can have a vector of booleans V. V[i]==true if M[E[i]]>0 and V[i]==false if M[E[i]]==0. So you begin by setting all the values of V at false, and each time you do M[S[R]]++, you can set V of this element to true, and each time you do M[S[L]]-- and M[S[L]]==0 then set V of this element to false