what does O(N) mean [duplicate]
The comment was referring to the Big-O Notation.
Briefly:
- O(1) means in constant time - independent of the number of items.
- O(N) means in proportion to the number of items.
- O(log N) means a time proportional to log(N)
Basically any 'O' notation means an operation will take time up to a maximum of k*f(N)
where:
k is a constant multiplier
f() is a function that depends on N
O(n) is Big O Notation and refers to the complexity of a given algorithm. n refers to the size of the input, in your case it's the number of items in your list.
O(n) means that your algorithm will take on the order of n operations to insert an item. e.g. looping through the list once (or a constant number of times such as twice or only looping through half).
O(1) means it takes a constant time, that it is not dependent on how many items are in the list.
O(n^2) means that for every insert, it takes n*n operations. i.e. 1 operation for 1 item, 4 operations for 2 items, 9 operations for 3 items. As you can see, O(n^2) algorithms become inefficient for handling large number of items.
For lists O(n) is not bad for insertion, but not the quickest. Also note that O(n/2) is considered as being the same as O(n) because they both grow at the same rate with n.
It's called Big O Notation: http://en.wikipedia.org/wiki/Big_O_notation
So saying that insertion is O(n) means that you have to walk through the whole list (or half of it -- big O notation ignores constant factors) to perform the insertion.
This looks like a nice introduction: http://rob-bell.net/2009/06/a-beginners-guide-to-big-o-notation/
Specifically O(n) means that if there's 2x as many items in the list, it'll takes No more than twice as long, if there's 50 times as many it'll take No more than 50 times as long. See the wikipedia article dreeves pointed out for more details
Edit (in bold above): It was pointed out that Big-O does represent the upper bound, so if there's twice as many elements in the list, insertion will take at most twice as long, and if there's 50 times as many elements, it would take at most 50 times as long.
If it was additionally Ω(n) (Big Omega of n) then it would take at least twice as long for a list that is twice as big. If your implementation is both O(n) and Ω(n), meaning that it'll take both at least and at most twice as long for a list twice as big, then it can be said to be Θ(n) (Big Theta of n), meaning that it'll take exactly twice as long if there are twice as many elements.
According to Wikipedia (and personal experience, being guilty of it myself) Big-O is often used where Big-Theta is what is meant. It would be technically correct to call your function O(n^n^n^n) because all Big-O says is that your function is no slower than that, but no one would actually say that other than to prove a point because it's not very useful and misleading information, despite it being technically accurate.