longest increasing/decreasing subsequence in O( n * lg n )

For this, you have to keep two variables which store indexes of the start and end element of the longest sequence found so far.

Strictly Increasing longest sequence--

lstart = lend = start = end = 0;
for (int i=0; i<size-1; i++)
{
  if (array[i+1] > array[i])
    end = i+1;
  else
    start = i+1;
  if (lend - lstart < end - start)
    { lstart = start; lend = end; }
}

kumvjuec's solution is O(n) and it works. Unfortunately it solves the wrong problem. It finds the longest increasing _sequential_ subsequence. That is, it finds the longest increasing run in the original array. When this problem is given in algorithms the problem is to find the longest subsequence where subsequence is a set of numbers from the original sequence IN THE SAME ORDER AS IN THE ORIGINAL but without regard to numbers BETWEEN them in the original sequence.

Kitty__Kong: In case there is any question, the same algorithm will solve both of your problems with the reversal of a single > or < sign.

What do you have so far? That is, what data structures are you using to support your solution? You can store the list of numbers as a vector<int> of the indices. To append a number to a list you just use push_back.

This is a fun problem with a lot of learning potential.

-bcl

int bs( vector< pair<int, int> > &L, int s, int f, int t )
{
      int      m;

      if( s > f )
      {
            while( s >= L.size() || L[s].first >= t )
                  s--;

            return s;
      }

      m = (s + f) / 2;

      if( L[m].first > t )
            return bs( L, s, m - 1, t );
      else if( L[m].first < t )
            return bs( L, m + 1, f, t );
      else
      {
            while( L[m].first == t )
                  m--;

            return m;
      }
}

vector<int> Proc( vector<int> &nums )
{
      int                                          i, a;
      vector< pair<int, int> >      L;
      vector<int>                              ret;

      //reverse( nums.begin(), nums.end() );

      L.reserve( nums.size() );

      L.push_back( pair<int, int>( INT_MIN, INT_MIN ) );

      for( i = 0; i < nums.size(); ++i )
      {
            a = bs( L, 0, L.size() - 1, nums[i] );

            if( a + 1 >= L.size() )
                  L.push_back( pair<int, int>( nums[i], L[L.size() - 1].first ) );
            else
            {
                  L[a + 1].first = nums[i];
                  //L[a + 1].second = L[a].first;
            }

            for( int j = 0; j < L.size(); ++j )
                  cout << L[j].first << " ";

            cout << endl;
      }

      /*L.pop_back();
      reverse( L.begin(), L.end() );*/

      L.erase( L.begin() );

      for( i = 1; i < L.size(); ++i )
            ret.push_back( L[i].second );

      ret.push_back( L[L.size() - 1].first );

      return ret;
}

this is what i have so far. the problem is i dont understand how to keep track of each numbers predecessor which im trying to do in the second member of the pair<int, int>

(1) Do you have notes on the data representation? I am assuming that someone gave it to you and you are trying to use it. A quick description of what L is supposed to contain (and why you are comparing the result of bs+1 to the length of L in Proc) would save me some head pounding.

-bcl

i found it using google

http://www.stanford.edu/class/cs161/Handouts/dpex.pdf

What predecessor are you keeping? That is, how are you encoding the sequence itself? L is related to the BEST subsequence of any given length. You will not be able to walk it to find the sequence (in particular, consider a sequence that ends with its second smallest value. In that instance L[2] will be replaced on the last iteration through the loop. Thus the predecessor of L[3] will have no predecessor available. I am pretty sure you're going to need more storage than they use in the agorithm you point to. Keeping each subsequence should not be too costly since you will only ever extend one by a single value. That would make L a vector<pair<int, vector<int> > instead of what you have.

I could be wrong. It has been a while since I taught this problem but I am pretty sure to get the actual sequence (rather than the length of the sequence) requires more storage.

-bcl

this is where im stuck. when i add an item to L how do i know what its predecessor is? i had tried using a vector<pair<int, vector<int> > but couldnt figure out what to insert in the vector in the pair.