Let $ε>0$. We consider the problem of constructing a Hamiltonian graph with $(1+ε)n$ 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 $(1+ε)n$ edges in $(1+ε)(1+(\log n)/2K) n$ rounds w.h.p. The case $K=1$ implies that w.h.p. one can build a Hamiltonian graph by choosing $(1+ε)n$ edges in an on-line fashion as they appear along the first $(0.5+ε)n\log n$ steps of the random graph process, this refutes a conjecture of Frieze, Krivelevich and Michaeli. The case $K=Θ(\log n)$ implies that the Hamiltonicity threshold of the corresponding Achlioptas process is at most $(1+ε)(1+(\log n)/2K) n$. This matches the $(1-ε)(1+(\log n)/2K) n$ lower bound due to Krivelevich, Lubetzky and Sudakov and resolves the problem of determining the Hamiltonicity threshold of the Achlioptas process with $K=Θ(\log n)$. 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 $(0.5+ε)n$ edges, within $(1+ε)(0.5+(\log n)/2K) n$ 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.