A non$-$recursive version of FIBONACCI$(n)$ is given below:

FIBONACCI $(n)$

if $n=0$

return $0$

else

if $n>0,//$the general case

let $F[0..n]$ be an array

$F[0]=0$

$F[1]=1$

for $i=2$ to $n$

$F[i]=F[i-1]+F[i-2]$

return $F[n]$

Explain convincingly why the time complexity of this version of FIBONACCI$(n)$ is $(n).$