What is the time complexity of following algorithm?

Question

Given a sequence of n integers, A. To find the sum of the sequence we are following the following recursive algorithm, we make a new sequence B of size n/2 where B[i] = A[2*i] + A[2*i + 1] for i from 0 to n/2 - 1 and we replace A with B. When size of A is 1, we return the element itself.

Shouldn't the time complexity be calcualted as follows?

T(n) = T(n/2) + O(n/2)
or
T(n) = T(n/4) + O(n/4) + O(n/2)
or
T(n) = O(1) + O(2) + O(4) + ... + O(n/4) + O(n/2)

At this point I am not sure if I am doing it correctly and what this value should be equal to. I am assuming O(nlgn)

How do I arrive at the solution ? Also, using master theorem is giving me O(n), I am not sure if I am applying master theorem properly. Can someone please guide me here?

It's not clear exactly what your "algorithm" is, but note that if summing a sequence takes longer than O(n), you're doing something wrong ;) — Oliver Charlesworth
– Oliver Charlesworth, Commented May 15, 2017 at 8:29

Dima Chubarov · Accepted Answer · 2017-05-15 08:41:25Z

5

First observe that

O(1) + O(2) + O(4) + ... + O(n/4) + O(n/2) =
  O(1 + 2 + 4 + ... + n/4 + n/2)

Then you could recognize here the sum of a geometric progression. Assuming n = 2^m (2 to the power of m)

  1 + 2 + 4 + ... + 2^(m-2) + 2^(m-1) = 2^m = n

Therefore your method gives you T(n) = O(n). This is in an agreement with the master theorem.

answered May 15, 2017 at 8:41

Dima Chubarov

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