# np complete – A variant of partition problem

I am thinking about the following variant version of partition problem:

Given $$n$$ positive integer numbers $$a_1, ldots, a_n$$ which sum to a even number $$S= sum_{i=1}^n a_i$$. The decision problem of the existence of a subset $$A$$ such that $$sum_{i in A}a_i = S/2 + 1$$ is NP-hard?

Here I’m essentially partitioning the ground set $${1,ldots, n}$$ into two sets whose summation differ by $$textbf{exactly}$$ 2. If this difference is $$textbf{at most}$$ 2 instead, then we can obtain a simple reduction from partition problem by multiplying every numbers by 3.