Question

I am given a question to write a method with a time complexity of O(4^n).

I have came out with this algorithm:

public void test(int n){
   for(int i = 0; i<n;i++){
      test(4*i);
   }
}

Is this be considered to be running on O(4^n)?

Answer 1

You were quite close with your program. The correct function would go like this:

public void test(int n) {
  if (n == 0) return;
  for (int i = 0; i < 4; i++) test(n-1);
}

Run this piece of code to check:

static int runs;
static void test(int n) {
  runs++;
  if (n == 0) return;
  for (int i = 0; i < 4; i++) test(n-1);
}
public static void main(String[] args) {
  for (int n = 1; n <= 5; n++) {
    runs = 0;
    test(n);
    System.out.format("%d: %d %d\n", n, 1<<(2*n), runs);
  }
}

It will print

1: 4 5
2: 16 21
3: 64 85
4: 256 341
5: 1024 1365

The run count is off by one, but the big-O complexity is satisfied.

The reasoning as to why it is O(4ⁿ) is probably quite obvious once you see it, but a little explanation cannot hurt. The function is best imagined to be solving a complex problem by divide-and-conquer. It reduces an n-sized problem to four instances of an (n-1)-sized problem, recursing until the subproblem is trivial (size 0). A problem of size 1 is therefore solved in 1+4 steps (entry-point call + 4 trivial subproblems); a problem of size 2 in 1 + 4*(1 + 4) = 21 steps, and so on.