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The optimum is

27 steps

because

the results yield 0 to 15
Press the following cells x amount of times:
\begin{matrix} 0 &1 &2 &1\\ 0 &7 &1 &1\\ 0 &2 &4 &6 \\ 0 &0 &1 &1 \end{matrix}

yielding

\begin{matrix} 1 &10 &5 &4\\ 7 &11 &15 &9\\ 2 &13 &14 &12 \\ 0 &3 &6 &8 \end{matrix}

I think the least amount of presses is

27

Press the following cells $x$ amount of times

\begin{matrix} 0 &1 &2 &1\\ 0 &7 &1 &1\\ 0 &2 &4 &6 \\ 0 &0 &1 &1 \end{matrix}

yielding

\begin{matrix} 1 &10 &5 &4\\ 7 &11 &15 &9\\ 2 &13 &14 &12 \\ 0 &3 &6 &8 \end{matrix}

Also not sure if this is optimal, but here's one with 28:

\begin{matrix} 0 &3 &4 &1\\ 0 &6 &4 &2\\ 0 &1 &2 & 3 \\ 0 &0 &1 &1 \end{matrix}

yielding

\begin{matrix} 3 &13 &12 &7\\ 6 &14 &18 &10\\ 1 &9 &11 &8 \\ 0 &2 &4 &5 \end{matrix}

The optimum is

27 steps

because

the results yield 0 to 15
Press the following cells x amount of times:
\begin{matrix} 0 &1 &2 &1\\ 0 &7 &1 &1\\ 0 &2 &4 &6 \\ 0 &0 &1 &1 \end{matrix}

yielding

\begin{matrix} 1 &10 &5 &4\\ 7 &11 &15 &9\\ 2 &13 &14 &12 \\ 0 &3 &6 &8 \end{matrix}

Also not sure if this is optimal, but here's one with 29:

\begin{matrix} 0 &3 &3 &1\\ 0 &1 &3 &0\\ 0 &2 &8 & 2 \\ 0 &0 &3 &3 \end{matrix}

yielding

\begin{matrix} 3 &7 &10 &4\\ 1 &9 &15 &6\\ 2 &11 &18 &13 \\ 0 &5 &14 &8 \end{matrix}

Also not sure if this is optimal, but here's one with 28:

\begin{matrix} 0 &3 &4 &1\\ 0 &6 &4 &2\\ 0 &1 &2 & 3 \\ 0 &0 &1 &1 \end{matrix}

yielding

\begin{matrix} 3 &13 &12 &7\\ 6 &14 &18 &10\\ 1 &9 &11 &8 \\ 0 &2 &4 &5 \end{matrix}