Minimum number of vertices to remove to bound the largest connected component of a graph

Question

I have this problem, maybe anybody could help.

Given a graph $G = (V, E)$ and an integer $k \geq 1$, find the minimum number $l$ of vertices to remove to make the largest connected component of $G \setminus \{v_1, \dots,v_l\}$ have at most $k$ vertices.

I wonder it this problem can be solved quickly (in polynomial time)?

(Related: finding the vertices when $k$ is small)

Answer 1

The optimization version of this problem is known as Graph Integrity, which is unfortunately NP-Hard problem:

The integrity of a graph G, denoted $I(G)$, is defined by: $I(G) = min\{ |S| + m(G − S) : S \subset V (G) \}$ where $m(G − S)$ denotes the maximum order of a component of $G − S$;