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###### Hakim Rachidi

38,489 Points# Shouldn't Merge Sort be O(n * log(n) + log(n))

When you think about the Merge Sort you split the set ( log(n) ) and then you join them together while sorting each group (n * log(n)). In sum that such be log(n) + n * log(n)

## 2 Answers

###### Clayton Perszyk

Treehouse Moderator 48,565 PointsNot sure, but this looks like a good explanation: https://softwareengineering.stackexchange.com/questions/297160/why-is-mergesort-olog-n

###### Paul Brubaker

14,265 PointsThe reason that O(n*log(n)+log(n)) simplifies to just O(n*log(n)) is that the limit as n approaches infinity of the ratio of n*log(n)+log(n) to n*log(n) is 1. In other words, as we make the size of the input set arbitrarily large, the rate of growth of n*log(n)+log(n) and n*log(n) is 'almost exactly' the same so we just write O(n*log(n)). I hope this helps, I don't think this course is intended to get into the weeds with calculus but limits aren't too hard to understand intuitively. Overlaying the graphs of nlog(n) and nlog(n) + log(n) for large n as in the video examples might help to illustrate the point.

## Hakim Rachidi

38,489 Points## Hakim Rachidi

38,489 PointsThanks, but the Stack post is not very helpful as the answers just explain what I already said. The set first get's split (log(n) operations) then concatenated while sorting again per split (n * log(n)). In total: n * log(n) + log(n).

Here is my post on SO