Consider the following C - function:

double foo(int n){
 int i;
 double sum;
 if(n == 0) return 1.0;
 sum = 0.0;
 for (i = 0; i < n; i++){
  sum += foo(i);
 }
 return sum;
}

Suppose we modify the above function foo() and store the values of foo(i), $$0 \le i \le n$$, as and when they are computed. With this modification, the time complexity for function foo() is significantly reduced. The space complexity of the modified function would be:

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

The time complexity of the following C function is (assume n > 0)

int recursive(int n){
 if(n == 1){
   return (1);
 }
 return (recursive(n - 1) + recursive(n - 1));
}

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