Subsets of a group with special property Let $G$ be a finite group. We say a subset $A$ of $G$, $|A|=m$, is $(m,i)$-good, $m\geq 1$ and $0\leq i\leq m$, if there exist $g_A\in G$ such that we have $|gA\cap A|=m-i$.
I need some groups such that, for fixed $m$ and $i$, preferably $i$ is much smaller than $m$, the number of $(m, i)$-good subsets is large.
Is this property well known in group theory? Or, is it related to some well-known properties of groups for special values $m$ and $i$? 
$\textbf{The motivation for this question:}$ These type of subsets are used to define some special type of participants in secret sharing schemes on groups. Participants are the members of group and subsets with this property can take some special values of share to construct the key.
I do not know how can I upload a large file here. You can download the text file from this address. If it is helpful, I can do this computation for each order of groups.
$\textbf{Addendum 1}:$ http://s8.picofile.com/file/8341377742/Order_8.txt.html
$\textbf{Addendum 2}:$ http://s9.picofile.com/file/8341380550/Order_9.txt.html
$\textbf{Addendum 3}:$ http://s9.picofile.com/file/8341379792/Order_12.txt.html
$\textbf{New question (10/2018)}:$
Is it true that if the $(m,i)$-good numbers of two groups $G$ and $H$ are equal, then $G$ is isomorphic to $H$?
Thanks for helpful answers, references, and comments.
 A: Any $m$-set is an $(m,0)$-set with $g$ the identity. It is easier to count the number of pairs $(g,S)$ where $S$ is $(m,i)-$good for that particular $g.$ If $g$ has order $2$ and $|G|=n=2t$ then, using that $g$, there are $\binom{t}k$ sets that are $(2k,0)$ and $\binom{t}k\binom{t-k}i2^i$ which are $(2k+i,i). $ An elementary abelian $2$ group has all non-identity elements of order $2$. Of course the same set could be $(m,i)$ for more than one $g$ but that might not be that big a consideration.
It doesn't really matter if the group is abelian. For $g$ of order $j,$ the exact counts using that $g$, are sums of terms each the product of a multinomial coefficient and some powers of constants which are counts for a cyclic group of order $j$.
For $C_s$ and a given generator $g$ consider the $2^s-1$ non-empty subsets and let $c_{m',i'}$ be the number that are $(m',i')-$good for that $g.$ 
Then if $G$ is a group of order $n=st$ and $g$ is an element of order $s$ then from the $t$ right cosets of $<g>$ $n_{(m',i')}$ of each type. The result will be $(m,i)-$ good (for that $g$) for $m=\sum n_{(m',i')}m'$ and $i=\sum n_{(m',i')}i'.$ Sum over all the possibilities to get the total count of $(m,i)$ sets for that $g$. The number of ways to do this is a multinomial coefficient.
To count $(m,i)-$good sets with a distinguished $g$ do the above for every combination giving the desired $(m,i)$
In theory, do the above for each element of $G$ then somehow account for sets with multiple possible $g$.
