So this is the Master theorem for Merge Sort:

$$ T(n) = 2T(n/2) + \Theta(n). $$

I am not able to understand why is the time complexity for sorting and merging $\Theta(n)$.

Is sorting $O(1)$ and merging $O(n)$?

  • 3
    $\begingroup$ No, that is not the "Master's theorem" for merge sort; that is simply its time complexity expressed as a recursion which can be asymptotically approximated using the master theorem (no capital and no apostrophe; "master" is not someone's name). $\endgroup$
    – dkaeae
    Commented Feb 20, 2019 at 9:50
  • 1
    $\begingroup$ Also, if you mean $\Theta(n)$ (instead of $\theta(n)$), merging takes worst-case linear steps since, in the worst-case, you merge 2 lists of (approximately) $\frac{n}{2}$ elements each. $\endgroup$
    – dkaeae
    Commented Feb 20, 2019 at 9:51

1 Answer 1


Each iteration of merge sort consist of 2 phases:

  1. Merge Sorting the first and the second half separately.
  2. Merging the two halves.

So in your equation phase 1 is represented by $2T(n/2)$. This means that merge sort is called on the two halves. This is a recursive call, which is why $T$ is used here.

Phase 2 is represented by $\Theta(n)$. Merging two lists of length $a$ and length $b$ takes $\Theta(a+b)$. In our case the two lists are both of size $\frac{1}{2}n$, so we get a total of $\Theta(n)$.

In the end we get that $T(n) = $ step 1 + step 2, resulting in $T(n) = 2T(n/2) + \Theta(n)$


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