  In the {\sc Min-Sum 2-Clustering} problem, we are given a graph and a
parameter $k$, and the goal is to determine if there exists a 2-partition of
the vertex set such that the total conflict number is at most $k$, where the
conflict number of a vertex is the number of its non-neighbors in the same
cluster and neighbors in the different cluster. The problem is equivalent to
{\sc 2-Cluster Editing} and {\sc 2-Correlation Clustering} with an additional
multiplicative factor two in the cost function. In this paper we show an
algorithm for {\sc Min-Sum 2-Clustering} with time complexity $O(n\cdot
2.619^{r/(1-4r/n)}+n^3)$, where $n$ is the number of vertices and $r=k/n$.
Particularly, the time complexity is $O^*(2.619^{k/n})$ for $k\in o(n^2)$ and
polynomial for $k\in O(n\log n)$, which implies that the problem can be solved
in subexponential time for $k\in o(n^2)$. We also design a parameterized
algorithm for a variant in which the cost is the sum of the squared
conflict-numbers. For $k\in o(n^3)$, the algorithm runs in subexponential
$O(n^3\cdot 5.171^{\theta})$ time, where $\theta=\sqrt{k/n}$.
