You asked about the center of a group with an efficient presentation: the relation of "central" to "efficient presentation" is probably strongest in the Schur multiplier. In particular, a perfect group can only have an efficient presentation if its Schur multiplier is trivial (so it is the universal perfect central extension of its inner automorphism group). If the inner automorphism group has a non-trivial Schur multiplier, then the perfect group with an efficient presentation must have a non-trivial center.
However, many perfect groups have trivial Schur multipliers and centers. For instance, the Mathieu group M11 has an efficient presentation on two generators with two relations, and of course has a trivial center.
Efficient presentations are somewhat poorly named (balanced might be better) since they are typically awful for computational group theory. However, people have taken them as a challenge, and so you can find scores of papers dealing with efficient presentations of (covering groups of) finite simple groups. One paper with nice tables is:
Campbell, Colin M.; Havas, George; Ramsay, Colin; Robertson, Edmund F.
Nice efficient presentations for all small simple groups and their covers.
LMS J. Comput. Math. 7 (2004), 266–283.
MR2118175
Online version.
Note that efficient is sometimes defined in terms of the rank of the Schur multiplier. So be careful that your usage may contradict some of the papers (which give "efficient" presentations for groups with non-trivial Schur multiplier; they are allowed to use one extra relation per independent generator of the Schur multiplier).
A related concept is an asymptotic version that seeks to describe how complicated presentations of finite simple groups get. In fact it seems they are very, very simple:
Guralnick, R. M.; Kantor, W. M.; Kassabov, M.; Lubotzky, A.
Presentations of finite simple groups: a quantitative approach.
J. Amer. Math. Soc. 21 (2008), no. 3, 711–774.
MR2393425
DOI:10.1090/S0894-0347-08-00590-0
where it is shown that there is an absolute constant C so that a total C generators and relations is needed to define a presentation of a finite quasi-simple group, and indeed the total word length of the relations is bounded by C*(log(n)+log(q)) where n is the rank of the BN-pair and q is the size of the field of definition (and alternating and sporadic groups just get less than C). They left out the case of Ree groups (the characteristic 3 kind), but probably it will be taken care of at some point.
If you want to explore these things in GAP, part of the documentation for GAP is dedicated to this problem:
http://www.gap-system.org/Doc/Examples/balanced.html
Note that $n$ is the sum over prime divisors $p$ of $|G|$ of the minimal number of generators of the distinct Sylow $p$-subgroups of $G.$ The sizes of all minimal generating sets of a finite $p$-group are the same by properties of the Frattini subgroup. Use of the Frattini subgroup helps to prove the leftmost inequality: take a prime $p$ which divides $d_1 .$ Then a Sylow $p$-subgroup of $G$ can't be generated by fewer than $k$ elements, so $G$ itself certainly can't be generated by fewer than $k$ elements, as each Sylow $p$-subgroup of $G$ is a homomorphic image of $G.$ On the other hand, take a minimal generating set $S$ for $G$ of maximal cardinality, and minimize the sum of the orders of elements of $S$ subject to that. Then each element of $S$ must have prime power order, for if $s \in S$ has order divisible by more than one prime, then we may write $s = t + u $ where $t$ and $u$ have coprime orders (each greater than one) whose product is the order of $s$. Then $(S \backslash \{ s \}) \cup \{t,u\}$ is still a minimal generating set for $G,$ contradicting the maximality of the cardinality of $S.$ The fact that $S$ is a minimal generating set means that if we now collect the elements of $S$ whose orders are powers of a fixed prime $p$, we must obtain a generating set for a Sylow $p$-subgroup of $G,$ and this must be minimal by the choice of $S$. Hence the cardinality of $S$ is at most $n,$ as defined above.
Best Answer
This is an answer for the abelian case only.
For a finite abelian group (let us exclude the trivial one) there are two standard ways to decompose it as a direct sum of cyclic groups. One into cyclic groups of order $n_1,\dots,n_r$ (not $1$) such that $n_i \mid n_{i+1}$ and the other one into cyclic groups of order $m_1, \dots, m_s$ such that each $m_i$ is a prime power (not $1$). [Under each of the assumptions 'divisibility' and 'prime power' the repective decomposition is unique, in the latter case of course up to the ordering.]
The parameter $r$ is sometimes called the rank and the parameter $s$ the total rank of the group (although terminology here is not completely uniform).
Now, it is known that the rank is the minimal cardinality of a generating set, in the sense that there does not exists a set of a smaller cardinality that generates the group. And, that the total rank is the maximal cardinality of a minimal generating set, that is there exists a generating set of cardinailty $s$ such that no subset of this set generates the group.
Thus, the cardinality of all minimal/irredundant generating set is uniquely determined if and only if the rank equals the total rank. This is the case if and only if the group is an (abelian) $p$-groups.
Note that for a non-$p$-group on can first consider the first decomposition into cycylic groups, and then decompose each cyclic component as the sum of cyclic groups of prime power order (more or less Chines Remainder Theorem); so only if all the orders in the first decomposition are also prime powers (and thus necessarily powers of the same prime) does one not get a different value for rank and total rank.