  Let $\epsilon>0$. We consider the problem of constructing a Hamiltonian graph
with $(1+\epsilon)n$ edges in the following controlled random graph process.
Starting with the empty graph on $[n]$, at each round a set of $K=K(n)$ edges
is presented, chosen uniformly at random from the missing ones (or from the
ones that have not been presented yet), and we are asked to choose at most one
of them and add it to the current graph. We show that in this process one can
build a Hamiltonian graph with at most $(1+\epsilon)n$ edges in
$(1+\epsilon)(1+(\log n)/2K) n$ rounds w.h.p. The case $K=1$ implies that
w.h.p. one can build a Hamiltonian graph by choosing $(1+\epsilon)n$ edges in
an on-line fashion as they appear along the first $(0.5+\epsilon)n\log n$ steps
of the random graph process, this refutes a conjecture of Frieze, Krivelevich
and Michaeli. The case $K=\Theta(\log n)$ implies that the Hamiltonicity
threshold of the corresponding Achlioptas process is at most
$(1+\epsilon)(1+(\log n)/2K) n$. This matches the $(1-\epsilon)(1+(\log n)/2K)
n$ lower bound due to Krivelevich, Lubetzky and Sudakov and resolves the
problem of determining the Hamiltonicity threshold of the Achlioptas process
with $K=\Theta(\log n)$.
  We also show that in the above process w.h.p. one can construct a graph $G$
that spans a matching of size $\lfloor V(G)/2) \rfloor$, with $(0.5+\epsilon)n$
edges, within $(1+\epsilon)(0.5+(\log n)/2K) n$ rounds.
  Our proof relies on a robust Hamiltonicity property of the strong $4$-core of
the binomial random graph which we use as a black-box. This property allows it
to absorb paths covering vertices outside the strong $4$-core into a cycle.
