I am currently learning about Big O Notation running times and amortized times. I understand the notion of O(n) linear time, meaning that the size of the input affects the growth of the algorithm proportionally...and the same goes for, for example, quadratic time O(n2) etc..even algorithms, such as permutation generators, with O(n!) tim...
Hi, I've been having some problems trying to grasp the concept of big O notation. So, by definition big O is as follows, T(n) ∈ O(G(n)) if T(n) <= G(n) * C. Since the the constant "C" can be any integer > 0, wouldn't this following example be true as well? Example: n log n ∈ O(log n) n log n <= log n * c Where C is equal to the va...
for f = n(log(n))^5 g = n^1.01 is f = O(g) f = 0(g) f = Omega(g)? I tried dividing both by n and i got f = log(n)^5 g = n^0.01 But I am still clueless to which one grows faster. Can someone help me with this and explain the reasoning to the answer? I really want to know how (without calculator) one can determine which one grow...