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 computers … Read more
If you get stuck here’s a solution import math baseInt = 2020 #This is the int you want to find within the other int randomInt = 1012020 #This is the int you’re going to count up to def checkBaseInt(baseInt, theIteratorInt): theNumString = str(theIteratorInt) #Converting to strings to find the substring baseIntString = str(baseInt) if(theNumString.find(baseIntString) == … Read more
The way I see it, your code would be O(n) if it wasn’t for the a.pop(0) part. Since lists are implemented as arrays in memory, removing an element at the top means that all elements in the array need to moved as well. So removing from a list is O(n). You do that in a … Read more
Almost all that this code do is organizing input and output except only this line: count=Collections.frequency(list,x); This is the only line where real computations happen. Actually, it’s time and space complexity is determined by the standard java.util.Collections.frequency method which should have pretty good time and space characteristics for such a trivial case. 1 solved How … Read more
Upto my knowledge, most common time complexity notation is Big-O, I’ll share same in this answer. Time complexity is O(1), with an assumption that Math.Pow calculation is O(1) and .ToString() method is O(1). It is O(1) because each step will be executed only once. In gerenal, if we take .ToString().Length time complexity into account then … Read more
You will have to use some mathematics here. Lets talk about your 2nd and 3rd loop: No of iterations in 3rd loop(for c) for each iteration of 2nd loop(for b): 1 = 2^0 ( for b < 1) 2 = 2^1 ( for b < 2) 4 = 2^2 ( for b < 4) . … Read more
Introduction Analysis complexity is an important concept in computer science, as it helps to determine the efficiency of algorithms. In particular, the analysis of complexity in a for inside for loop is important, as the complexity of the second for loop depends on the first one. This article will discuss the analysis of complexity in … Read more
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). solved Giving the Big O, Big Theta and Big Omega for a function [closed]
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. In … Read more
n is large while r is small, you are better off compute nCr by n(n-1)…(n-r+1)/(1*2*…*r) You may need to find multiplicate inverse of 1, 2, … r mod 10^9+7 solved nCr mod 10^9 + 7 for n
List<Type> list = … list.removeIf(item -> item.getId() == something); As the name suggests, removeIf() removes all elements if they satisfy the predicate. 7 solved How to remove some items from an arraylist without loop and less time complexity?
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 : arr) … Read more
First, let’s look at two simple nested loops first: for i (1..N) { for j (1..N) { f(); } } for i (1..N) { for j (i..N) { g(); } } f() is called N*N = N2 = O(N2) times. g() is called N+(N-1)+…+5+4+3+2+1 = N(N+1)/2 = N2/2 + N/2 = O(N2) times. As you … Read more
What is the time complexity of the following code snippet – Assume x is a global variable and “statement” takes O(n) time? solved What is the time complexity of the following code snippet – Assume x is a global variable and “statement” takes O(n) time?
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 swapping … Read more