Firstly, you have not measured the speed of len(), you have measured the speed of creating a list/set together with the speed of len(). Use the –setup argument of timeit: $ python -m timeit –setup “a=[1,2,3,4,5,6,7,8,9,10]” “len(a)” 10000000 loops, best of 3: 0.0369 usec per loop $ python -m timeit –setup “a={1,2,3,4,5,6,7,8,9,10}” “len(a)” 10000000 loops, … Read more
One of the things you have to realise when dealing with O notation is that it is often very important to understand what n is. If the n is something related to something you can control (eg: the number of elements in a list you want sorted) then it makes sense to look hard at … Read more
The main purpose of the Big-O notation is to let you do the estimates like the ones you did in your post, and decide for yourself if spending your effort coding a typically more advanced algorithm is worth the additional CPU cycles that you are going to buy with that code improvement. Depending on the … Read more
It’s O(1). You can google for the source code and you will come to such: From http://www.docjar.com/html/api/java/util/LinkedList.java.html All of the Collection classes I have looked at store a the size as a variable and don’t iterate through everything to get it.
It appears to be O(n) in V8 (chrome, node.js) at least: > var hash = {} > , c = 0; > > var s = +new Date();Object.keys(hash);console.log(+new Date() – s); 0 > for(var i=0; i<100000; i++, c++){ hash[c] = 1; } > var s = +new Date();Object.keys(hash);console.log(+new Date() – s); 26 > for(var i=0; … Read more
Generate all the permutations of a list You have n! lists, so you cannot achieve better efficiency than O(n!).
O(n!) isn’t equivalent to O(n^n). It is asymptotically less than O(n^n). O(log(n!)) is equal to O(n log(n)). Here is one way to prove that: Note that by using the log rule log(mn) = log(m) + log(n) we can see that: log(n!) = log(n*(n-1)*…2*1) = log(n) + log(n-1) + … log(2) + log(1) Proof that O(log(n!)) … Read more
Yes, Radix Sort and Counting Sort are O(N). They are NOT comparison-based sorts, which have been proven to have Ω(N log N) lower bound. To be precise, Radix Sort is O(kN), where k is the number of digits in the values to be sorted. Counting Sort is O(N + k), where k is the range … Read more
Asymptotic complexity (which is what both big-O and big-Theta represent) completely ignores the constant factors involved – it’s only intended to give an indication of how running time will change as the size of the input gets larger. So it’s certainly possible that an Θ(n) algorithm can take longer than an Θ(n2) one for some … Read more
It turned out, I misunderstood Logn to be lesser than 1. As I asked few of my seniors i got to know this today itself, that if the value of n is large, (which it usually is, when we are considering Big O ie worst case), logn can be greater than 1. So yeah, O(1) … Read more