You model the time function to calculate Fib(n) as sum of time to calculate Fib(n-1) plus the time to calculate Fib(n-2) plus the time to add them together (O(1)). This is assuming that repeated evaluations of the same Fib(n) take the same time – i.e. no memoization is used. T(n<=1) = O(1) T(n) = T(n-1) … 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
The first code sample is pretty much a classis for loop. Its complexity is O(n^2). This is because the inner loop has a complexity O(n) and it is run n times. The second one is a bit more difficult, untill you see that is equivalent to a non nested loop (ignoring the complexity of the … Read more
Any function whose runtime has the form (log n)k is O((log n)k). This expression isn’t reducable to any other primitive function using simple transformations, and it’s fairly common to see algorithms with runtimes like O(n (log n)2). Functions with this growth rate are called polylogarithmic. By the way, typically (log n)k is written as logk n, so the above … Read more
f ∈ O(g) says, essentially For at least one choice of a constant k > 0, you can find a constant a such that the inequality 0 <= f(x) <= k g(x) holds for all x > a. Note that O(g) is the set of all functions for which this condition holds. f ∈ o(g) … Read more
Explanation: The outer for loop should be clear; it is executed n times. Now to the inner loop. In the inner loop, you take n and always divide it by 2. So, you ask yourself: How many times can I divide n by 2? It turns out that this is O (log n). In fact, the base of log is 2, but in Big-O notation, we remove … Read more
EDIT: I’ve revised my answer based on the comments I’ve got this HW question which asks me to order a list of functions by their growth rate. The question also asks to indicate which ones have the same growth rate. Here are the functions: N sqrt(N) N^1.5 N^2 N log N N log log N … Read more
Just ask wolframalpha if you have doubts. In this case, it says Or you can also calculate the limit yourself: That means n^2 grows faster, so n log(n) is smaller (better), when n is high enough.
I can see how, when looking up a value in a BST we leave half the tree everytime we compare a node with the value we are looking for. However I fail to see why the time complexity is O(log(n)). So, my question is: If we have a tree of N elements, why the time … Read more
When you gotta do something n times, you have to do it n times. You can use a for loop, a while loop or whatever that your programming language (or pseudo code!) offers. Crudely, big O notation comments on the amount of work you have to do, and doesn’t care about how you do it … Read more