Yes, nested loops are one way to quickly get a big O notation. Typically (but not always) one loop nested in another will cause O(n²). Think about it, the inner loop is executed i times, for each value of i. The outer loop is executed n times. thus you see a pattern of execution like this: … Read more
As @emory pointed out, it is provably impossible to determine the big-O time complexity of an arbitrary piece of code automatically (the proof is a reduction from the halting problem). However, there are tools that can attempt to measure the complexity of a piece of code empirically by running it on several different inputs. One … Read more
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
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
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
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
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
n log(n) implies that the algorithm looks at all N items log(n) times. But that’s not what’s happening with Quickselect. Let’s say you’re using Quickselect to pick the top 8 items in a list of 128. And by some miracle of random selection, the pivots you pick are always at the halfway point. On the … 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.