Longest subsequence such that A[i].x < A[i+1].y

Question

I have an issue for which I am looking for an algorithm (if it exists)

What I have: An array of items which have certain properties, e.g. item $A$ has properties $x$ and $y$.

Example: $[ A(x,y), B(x,y), C(x,y), D(x,y), E(x,y) ]$

What I want: A result list consisting of elements of the original list, such as $[ A(x,y), C(x,y), E(x,y) ]$, for which the following properties are true:

• No reordering of elements, they are in the same order as the original list
• The result has the maximum number of elements, i.e. the longest 'path' possible
• For each pair of consecutive items $(A(x,y), B(x,y))$ in the result, $A.x \lt B.y$. In other words, an item's $x$ must be less than the next item's $y$.

Complexity: The list in the case I have is about 35 items long, so an algorithm which is $O(n!)$ might not work.

Does such an algorithm exist?

Answer 1

It's easier to change notation. Suppose the array is $A$, with the $i$-th element denoted $A[i]$ for $i=1,2,\dots,n$, and that element $i$ has attributes $A[i].x$ and $A[i].y$. Associate a directed graph $G$ with the array as follows. The vertices of $G$ are the indices $1\dots n$ of the array. Vertex $i$ is connected to vertex $j\ $ if both conditions $i<j$ and $A[i].x < A[j].y\ $ hold.

You are then asking for a longest directed path in $G$.

Note that $G$ is actually acyclic, since the arcs form a subset of the usual order on the set $\{1,\dots,n\}$. Any standard linear algorithm can then be used to compute the longest path, such as a topological sort.

Answer 2

So, your comparison is based on numerical properties x and y? If this is the case, then do some kind of sorting algorithm first, and adjust the order after sorting so that your condition of no reordering holds (that is delete those elements that are not at the correct location).