Computational complexity of least square regression operation In a least square regression algorithm, I have to do the following operations to compute regression coefficients:


*
  
*Matrix multiplication, complexity: $O(C^2N)$
  
*Matrix inversion, complexity: $O(C^3)$
  
*Matrix multiplication, complexity: $O(C^2N)$
  
*Matrix multiplication, complexity: $O(CN)$


where, N are the training examples and C is total number of features/variables.
How can I determine the overall computational complexity of this algorithm?
EDIT:
I studied least square regression from the book Introduction to Data Mining by Pang Ning Tan. The explanation about linear least square regression is available in the appendix, where a solution by the use of normal equation is provided (something of the form $a=(X^TX)^{-1}X^Ty)$, which involves 3 matrix multiplications and 1 matrix inversion). 
My goal is to determine the overall computational complexity of the algorithm.
Above, I have listed the 4 operations needed to compute the regression coefficients with their own complexity. Based on this information, can we determine the overall complexity of the algorithm?
Thanks!
 A: For a least squares regression with $N$ training examples and $C$ features, it takes:


*

*$O(C^2N)$ to multiply $X^T$ by $X$

*$O(CN)$ to multiply $X^T$ by $Y$

*$O(C^3)$ to compute the LU (or Cholesky) factorization of $X^TX$ and use that to compute the product $(X^TX)^{-1} (X^TY)$


Asymptotically, $O(C^2N)$ dominates $O(CN)$ so we can forget the $O(CN)$ part.
Since you're using the normal equation I will assume that $N>C$ - otherwise the matrix $X^T X$ would be singular (and hence non-invertible), which means that $O(C^2N)$ asymptotically dominates $O(C^3)$.
Therefore the total time complexity is $O(C^2N)$. You should note that this is only the asymptotic complexity - in particular, for $C$, $N$ smallish you may find that computing the LU or Cholesky decomposition of $X^T X$ takes significantly longer than multiplying $X^T$ by $X$.
Edit: Note that if you have two datasets with the same features, labeled 1 and 2, and you have already computed $X_1^T X_1$, $X_1^TY_1$ and $X_2^TX_2$, $X_2^TY_2$ then training the combined dataset is only $O(C^3)$ since you just add the relevant matrices and vectors, and then solve a $C\times C$ linear system.
In particular this allows you do to very fast bootstrap, jackknife and cross-validation when you are training an OLS regression (or variants like ridge regression, lasso, constrained OLS etc).
A: I was confused at why the matrix multiplication was claimed to be $\mathcal{O}(C^2 N)$ here, which is $\mathcal{O}(N^3)$ when $N=C$, which is much slower than Strassen's $\mathcal{O}(N^{2.8})$ and Le Gall's $\mathcal{O}(N^{2.37})$. 
It turns out that we are doing rectangular matrix multiplication where $N \ggg C$. There's much more data than variables. 
There are algorithms with far better scaling than the naive $\mathcal{O}(C^2N)$ cost for the dominant part in Chris Taylor's answer. 
You can multiply $X^TX$ with complexity $\mathcal{O}(C^{1.8}N)$ if using Strassen's $\mathcal{O}(C^{2.8})$ cost for $C \times C$ matrices, and in theory you can also do it with $\mathcal{O}(C^{1.37}N)$ if using the best known complexity for multiplication of $C \times C$ matrices (but keep in mind that Strassen's algorithm is often preferred because of the big constant in the latter algorithm). 
I obtained this from the first equation of page 262 of this paper, by setting:


*

*$m=p=C$

*$n=N$

*$N > C$

*$\omega = 2.8$ for Strassen, and $2.37$ for the algorithm with best theoretical (but not practical) asymptotic cost.

A: In this work https://www.research-collection.ethz.ch/bitstream/handle/20.500.11850/153646/eth-6011-01.pdf?sequence=1&isAllowed=y two implementation possibilities (the Gaussian elimination alternative vs. using the QR decomposition) are discussed in pages 32 and 33 if you are interested in the actual cost DFLOP-wise.
