计算算法的T（n）时间复杂度

Question

I am looking for some clarification in working out the time efficiency of an Algorithm, specifically T(n). The algorithm below is not as efficient as it could be, though it's a good example to learn from I believe. I would appreciate a line-by-line confirmation of the sum of operations in the code:

Pseudo-code

 1.  Input: array X of size n
 2.  Let A = an empty array of size n
 3.  For i = 0 to n-1
 4.      Let s = x[0]
 5.      For j = 0 to i
 6.          Let sum = sum + x[j]
 7.      End For
 8.      Let A[i] = sum / (i+1)
 9.  End For
 10. Output: Array A

My attempt at calculating T(n)

 1.  1
 2.  n
 3.  n
 4.  n(2)
 5.  n(n-1)
 6.  n(5n)
 7.  -
 8.  n(6)
 9.  -
 10. 1

 T(n) = 1 + n + n + 2n + n^2 - n + 5n^2 + 6n + 1
      = 6n^2 + 9n + 2 

So, T(n) = 6n^2 + 9n + 2 is what I arrive at, from this I derive Big-O of O(n^2). What errors, if any have I made in my calculation...

Edit: ...in counting the primitive operations to derive T(n)?

Answer 1

Your result O(n^2) is correct and is given by the two nested loops. I would prefer the derivation like

0 + 1 + 2 +  + (n-1) = (n-1)n/2 = O(n^2)

that follows from observing the nested loops.

Answer 2

I'm not really sure on your methodology but O(n^2) does seem to be correct. At each iteration through the first loop you do a sub loop of the previous elements. Therefore you're looking at 1 the first time 2 the second then 3 then... then n the final time. This is equivalent to the sum from 1 to n which gives you complexity of n^2.