Assuming the original instance is not easy in NP-hardness reduction
A polynomial time reduction from PARTITION to a language $L$ is a polynomial time function $f$ such that if $x$ is a Yes instance of PARTITION then $f(x) \in L$, and if $x$ is a No instance of PARTITION then $f(x) \notin L$.
If $x$ is an &quot;easy&quot; instance — belongs to some polynomially solvable class of instances — then $f$ can just solve it, outputting a fixed Yes or No instance of $L$. Therefore you can assume that $x$ is a &quot;hard&quot; instance.