Let A[1,...,n] be an array storing a bit (1 or 0) at each location, and f(m) is a function whose time complexity is O(m). Consider the following program fragment written in a C like language:

counter = 0;
for(i = 1; i <= n; i++){
 if(A[i]==1){
   counter++;
 }else{
   f(counter); counter = 0;
 }
}

The complexity of this program fragment is

What does the following algorithm approximate?
(Assume m > 1, $$ \in > 0$$)

x = m;
y = 1;
while(x - y > ε){
 x = (x + y) / 2;
 y = m/x;
}
print(x);

The recurrence equation
T(1) = 1
T(n) = 2T(n - 1)+n, $$n \ge 2$$
Evaluates to

The cube root of a natural number n is defined as the largest natural number m such that $${m^3} \le n$$. The complexity of computing the cube root of n (n is represented in binary notation) is