Time complexity for two pieces of code

Question

We've got 2 pieces of code:

int a = 3; 
while (a <= n) {
    a = a * a;
}

And:

public void foo(int n, int m) { 
    int i = m; 
    while (i > 100) 
        i = i / 3; 
    for (int k = i ; k >= 0; k--) { 
        for (int j = 1; j < n; j*=2) 
            System.out.print(k + "\t" + j); 
        System.out.println(); 
    } 
}

What is the time complexity of them? I think that the first one is: O(logn), because it's progressing to N with power of 2.
So maybe it's O(log²n) ?

And the second one I believe is: O(nlog2n), because it's progressing with jumps of 2, and also running on the outer loop.

Am I right?

The important thing is not looking at the data, but looking at the number of steps the algorithm needs. If the input doubles, how does the number of steps scale? Same? Double? Other? — Thorbjørn Ravn Andersen, Dec 26 '13 at 10:30

Mykhailo Granik · Accepted Answer · 2013-12-26T10:59:03.860

I believe, that first code will run in O(Log(LogN)) time. It's simple to understand in this way

Before first iteration you have 3 in power 1
After first iteration you have 3 in power 2
After second iteration you have 3 in power 4
After third iteration you have 3 in power 8
After fourth iteration you have 3 in power 16 and so on.

In the second code first piece of code will work in O(LogM) time, because you divide i by 3 every time. The second piece of code C times (C equals 100 in your case) will perform O(LogN) operations, because you multiply j by 2 every time, so it runs in O(CLogN), and you have complexity O(LogM + CLogN)

While your answer might be correct, without clarification it's useless, IMHO. — Maroun, Dec 26 '13 at 10:46

Martijn Courteaux · Answer 2 · 2013-12-26T11:43:30.743

4

For the first one, it is indeed O(log(log(n))). Thanks to @MarounMaroun for the hint, I could find this:

l(k) = l(k-1)^2
l(0) = 3

Solving this system yields:

l(k) = 3^(2^k)

So, we are looking for such a k that satisfies l(k) = n. So simply solve that:

enter image description here

This means we found:

enter image description here

The second code is seems misleading. It looks like O(nlog(n)), but the outer loop limited to 100. So, if m < 100, then it obviously is O(mlog(n)). Otherwise, it kind of depends on where exactly m is. Consider these two:

m: 305 -> 101 -> 33
m: 300 -> 100

In the first case, the outer loop would run 33 times. Whereas the second case would cause 100 iterations. I'm not sure, but I think you can write this as being O(log(n)).

edited Dec 26 '13 at 11:43

answered Dec 26 '13 at 10:39

Martijn Courteaux

67,591
47
198
287

1

I think the first loop is O(loglog(n)) – Maroun Dec 26 '13 at 10:39
I'll delete it from my answer and rethink in the meanwhile. – Martijn Courteaux Dec 26 '13 at 10:41
1

The first loop will loop until `a <= n`. So we're looking for `k` such that `2^2^k <= n`. When I see this, my instincts tell me "log of log". – Maroun Dec 26 '13 at 10:45
1

@MarounMaroun: Yes that is what I thought as well, but I can't really find a good explanation for it, other than "my instincts" :) – Martijn Courteaux Dec 26 '13 at 11:04
1

Well.. direct solving of the equation will yield that result, "proving" my instincts :)) – Maroun Dec 26 '13 at 11:06
@MarounMaroun: I updated my answer using your technique. Thank you for teaching me a new insight. – Martijn Courteaux Dec 26 '13 at 11:43
1

Now that's an answer.. too bad I can't +1 more than once. I think this answer will be helpful to many users, should gain more views :) – Maroun Dec 26 '13 at 11:45

Time complexity for two pieces of code

2 Answers2

Linked