A Question on HOD, V and GCH

The theorem 1.1 of the following paper:

Mohammad Golshani, V, HOD, and the GCH, Journal of Symbolic Logic.

states that:

Theorem: Assume $$V\models ZFC+GCH+~\text{There exists a}~(\kappa+4)-\text{strong cardinal}~\kappa$$ then there is a generic extension $$W$$ of $$V$$ such that:

(1) $$\kappa$$ remains inaccessible in $$W$$.

(2) $$V_{\kappa}^W=W_\kappa\models ZFC+\forall \lambda, ~2^{\lambda}=\lambda^{+3}$$.

(3) $$HOD^{W_{\kappa}}\models GCH$$.

The theorem shows the consistency of having a model of $$ZFC$$ (like $$W_{\kappa}$$) in which $$GCH$$ fails everywhere (in a finite gap form) but in its $$HOD$$, $$GCH$$ holds.

In a remark after the theorem the author says:

Remark: In fact it suffices to have a Mitchell increasing $$\kappa^+$$-sequence of extenders, each of which is $$(\kappa + 3)$$-strong. Thus for the conclusion of Theorem 1.1 it suffices to have a cardinal $$κ$$ such that $$o(κ) = \kappa^{+3}+\kappa^+$$.

My question is:

Question: Is $$o(κ) = \kappa^{+3}+\kappa^+$$ necessary for the proof of theorem 1.1. or one could hopefully reduce it to something weaker?

I guess here some sort of core model argument is needed to prove such a result but maybe there are other methods to ensure that the assumption could not be weaker.

1) In the above paper, we have a fixed gap 3 everywhere, while if someone wants a model in which $GCH$ fails everywhere but its $HOD$ satisfies the $GCH$, then we can have also gap 2. For this we need a measurable with $o(\kappa)=\kappa^{++}+\kappa^+,$ which is weaker than the assumption of the paper. With this assumption the arguments of the above paper work with few changes.
2) By a work of Gitik-Merimovich which is in preparation (see my answer given in Failure of the GCH), to get the total failure of $GCH$ we need large cardinals weaker than what I said above. Now two options may happen:
(A) It might be the case that their forcing is enough homogeneous so that the arguments of the above paper can be applied so that we have $GCH$ in its $HOD$, and then we will have an exact equiconsistency result,