[ad_1] I’ll do my best to explain it here on simple terms, but be warned that this topic takes my students a couple of months to finally grasp. You can find more information on the Chapter 2 of the Data Structures and Algorithms in Java book. There is no mechanical procedure that can be used … Read more
[ad_1] Assuming that you can insert integers of any size, the complexity will be the same as the complexity of division. So it is O(log n) due to log n is the number of digits. Notice 1: If you can only insert 32 or 64 bit integers, the “complexity” will be O(1). Notice 2: Since … Read more
[ad_1] n^k = (e^k)^log n and the factorial grows faster than an exponential (product of growing factors vs. product of constant factors). 1 [ad_2] solved Prove that (log n)! = O(n^k)
[ad_1] We have: 1 : if statement 3 : * operator 3 : function statement Totally: 7 So we have: T(n)=t(n-1) + t(n-2) + t(n-3) + 7 T(99)=1 T(98)=1 … T(1)=1 Then to reduce T(n), we have T(n-3)<T(n-2)<T(n-1) therefore: T(n) <= T(n-1)+T(n-1)+T(n-1) + 7 T(n) <= 3T(n-1) + 7 So we can solve T(n) = … Read more
[ad_1] If your a double loop that solves this problem breaks out of the loops as soon as a match is found, then both methods are similarly inefficient – they both require iterating over all of N elements M times, worst-case, where N is the length of one array and M is the length of … Read more
[ad_1] I’m not really sure why you want this, but you could do it with two nested loops: int getSum(int n) { int sum = 0; for(int i = 1; i <= n; i++) { int x = 0; while(x++ < i) { sum++; } } return sum; } This runs 1+2+3+…+n times, which simplifies … Read more
[ad_1] Drop all lower-order terms and constants and you get: Θ(T(n)) = Θ(n + 10*log(n)) = Θ(n) Since this is a tight bound (Θ) we also infer upper and lower bounds as O(n) and Ω(n). [ad_2] solved Giving the Big O, Big Theta and Big Omega for a function [closed]
[ad_1] It would be O(N). The general approach to doing complexity analysis is to look for all significant constant-time executable statements or expressions (like comparisons, arithmetic operations, method calls, assignments) and work out an algebraic formula giving the number of times that they occur. Then reduce that formula to the equivalent big O complexity class. … Read more
[ad_1] Clearly, O(n) time-complexity, as i iterates from 0 to n-1. [ad_2] solved What is the running time of this algorithm? [closed]
[ad_1] You are making your code way too complicated, use a HashSet<String>, which will guarantee uniqueness and will return whether the element was already in the set. public class DuplicateEle { public static void main(String args[]) { Set<String> seen = new HashSet<>(); String[] arr = { “hello”, “hi”, “hello”, “howru” }; for (String word : … Read more
[ad_1] list.reverse operates in place. That means that it almost certainly does something like this under the hood: for i in range(len(self) // 2): j = len(self) – 1 – i self[i], self[j] = self[j], self[i] The actual method is implemented in C, so of course it will be much more verbose, especially in the … Read more
[ad_1] 3n + 5n^2 + 1 | applying Big-O notation for the single terms = O(3n) + O(5n^2) + O(1) | leaving out constant multipliers = O(n) + O(n^2) + O(1) | taking the one with the maximum power = O(n^2) => quadratic time So you are right, the first option is the quadratic one. … Read more
[ad_1] To prove that f(n) = Θ(g(n)), you need to prove that there is a nonzero k such that lim (n → ∞) f(n) / g(n) = k In your case, we want to show that lim (n → ∞) lg(n + a) / (lg n) This is an indeterminate form of type ∞ / … Read more
[ad_1] The number of iterations of the while-loop is exactly floor(x/y). Each iteration takes n operations, because that is the complexity of the subtraction r – y. Hence the complexity of the algorithm is n * floor(x/y). However, we want to express the complexity as a function of n, not as a function of x … Read more
[ad_1] Is this a class assignment? I have a feeling you’re asking us to do your homework for you. The short answer is that the algorithm is O(n^3). The value of n is read from the console. The outer loop is i = 1 .. n (i is only ever incremented in the loop, i.e. … Read more