How do I find the worst-case time-complexity of an algorithm with an inner loop?

Question

Suppose I have a code with two for-loops:

int sum = 0;
for (int i = 1; i < N; i *= 2)
   for(int j = 0; j < i; j++)
      sum++;

How would I find the worst-case time complexity of this code? I've looked at many tutorials on finding time complexity and I understood them. But this one just seems a bit different from those in the tutorials.

Can you craft a `N` that is harder than any other `N`? Not all algorithms have a worst case.. — litelite, Sep 06 '17 at 18:27

score 3 · Accepted Answer · answered Sep 06 '17 at 18:32

3

Lets do simple maths. Value of i in each iteration goes as 1,2,4,8... (log N + 1) terms. In each iteration, the inner loop goes i times. Adding up the values of i

T(N) = 1 + 2 + ... + (log N + 1) terms i.e. GP with a = 1 and r = 2 and n = (log N + 1)

T(N) = a[(rⁿ-1)/(r-1)]
= 1[(2^(logN+1) - 1)/(2-1)]
= 2N - 1
= O(N)

So the complexity is O(N) in all scenarios.

answered Sep 06 '17 at 18:32

Shashwat Kumar

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Are you sure the last term in the sequence is logN + 1? Let's say N=100. We now that the value of i can't go higher than N (which is 100 in this case). So that maximum value of i is 64. But if we plug in 100 into logN + 1, we get 3. – TheSaviour Sep 06 '17 at 18:38
logN + 1 is terms in sequence, not the value of last term. Also, its log to the base 2. – Shashwat Kumar Sep 06 '17 at 18:44
I see. So if the inner loop was for `(int j = 0; j < N; j++)` instead, the complexity would still be O(N)? – TheSaviour Sep 06 '17 at 18:50
it's even simpler. why don't u try? – Shashwat Kumar Sep 06 '17 at 18:53
1

Here goes: The outer loop iterates (logN + 1) times and the j iterates N times per iteration of i. So I guess I should multiply N*(logN + 1)? So the complexity is O(NlogN)? – TheSaviour Sep 06 '17 at 18:59
Did i get it right or did I just embarrass myself? – TheSaviour Sep 06 '17 at 19:04
Yup... you got it :) – Shashwat Kumar Sep 06 '17 at 19:28
You can accept it if it solves your query. – Shashwat Kumar Sep 13 '17 at 17:03

Dima Chubarov · Answer 2 · 2017-09-06T18:47:40.220

1

First consider a simpler case when N=2^k+1 for some integer k. In this case the outer loop has k iterations and the overall number of operations is

 T(N) = 1 + 2 + 4 + ... + 2^k = 2^(k+1) - 1 = 2N - 3.

So the worst case complexity is at least O(2N - 3) = O(N).

Now suppose 2^k + 1 < N <= 2^(k+1) for some k. Then

 T(N) = 1 + 2 + 4 + ... + 2^k = 2^(k+1) - 1 < 2N = O(N)

Therefore T(N) = O(N).

edited Sep 06 '17 at 18:47

answered Sep 06 '17 at 18:27

Dima Chubarov

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How do I find the worst-case time-complexity of an algorithm with an inner loop?

2 Answers2