Consider the following C function.int fun1(int n) { int i,j,k,p,q=0; for (i=1; i<n; ++i) { p=0; for (j=n; j>1; j=j/2) ++p; for (k=1; k<p; k=k*2) ++q; } return q; }Which one of the following most closely approximates the return value of the function fun1?

Let us analyze the loops in fun1(int n):

1. Outer loop:for (i = 1; i < n; ++i) runs $(n-1)$ times, i.e., $\Theta(n)$ iterations.

2. First inner loop:for (j = n; j > 1; j = j / 2) ++p;

   • Variable $j$ is halved in each step starting from $n$ down to $1$.

   • Thus, $p$ increments $\lfloor \log_2 n \rfloor$ times, so after this loop, $p = \Theta(\log n)$.

3. Second inner loop:for (k = 1; k < p; k = k * 2) ++q;

   • Variable $k$ doubles in each step starting from $1$ until it reaches $p$.

   • Thus, the loop runs $\lfloor \log_2 p \rfloor = \lfloor \log_2 (\log_2 n) \rfloor = \Theta(\log(\log n))$ times.

   • In each iteration of this loop, $q$ is incremented by 1.

4. Since the outer loop executes $n-1$ times, the total increments to $q$ across all iterations are:

   $$q = (n-1) \times \Theta(\log(\log n)) = \Theta(n \log(\log n))$$

Therefore, the return value of fun1 most closely approximates $n\log(\log n)$, which corresponds to option (D).