As i is multiplied by 2 each time in a while, Hence the while loop will be run log(n) times. And inner for loop will be run in O(n^2) as i is at most n. Hence, the time complexity of the code in O notation is O(n^2 log(n)). solved What is the big O notation … Read more
Let’s consider your example… A = (5, 5, 1, 2, 3, 4, 5) If you consider the first element – 5 – then we know any solution must have one non-5 value at the other end, so we work backwards to find the first non-five, keeping track of the “front” and “back” locations we’re working … Read more
I wonder what can I do to increase the speed of the code. Can I get rid of some loops etc? The time complexity of each test case will be O(n) regardless of what you do, since you have to read all the numbers before you know when you stop printing zeros. Now, as for … Read more
So basically you have two loops, the top one is quite simple, but the second one is a tad bit more tricky. for (i = n/2; i < n; i) // ~n/2 so O(n) for(j = 0; j < n; j = 2 * j) // How many times does this run for n So … Read more
The body of the inner loop is executed n times. The body of the outer loop is executed for all powers of 2 from 1 to n and there are about log2(n) of them, hence the global complexity O(n.log(n)) 1 solved Showing the order of growth of the function
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 and … Read more
For positive ? the number of executions of count = count + 1 is representative of the complexity, as the other lines of code don’t execute significantly more times. The recurrence relation can use the first iteration of the outer loop: that represents ? iterations of the inner loop. The loop continues for ?/2, and … Read more
If finding a count of how many times a specific digit appears in a number is what you mean, than the complexity will be O(n) where n is the length of the string (number): char x = ‘7’; std::string number(“9876541231654810654984431”); int count = 0; for (size_t i = 0; i < number.size(); ++i) if (number[i] … Read more
A different argument: Your algorithm recurses until it “bottoms out” at fib(1) which returns 1. So you are getting the fibonacci number by doing (fibonacci number) calls and adding the 1s. So fib(n) is O(fib(n)). Binet’s Fibonacci formula can be written as fib(n) == int(0.5 + (phi ** n) / (5 ** 0.5)) where phi … Read more
We can list out a few terms T(0) = 0 T(1) = 1 * 0 + 1 T(2) = 2 * 1 + 2 = 4 T(3) = 3 * 4 + 3 = 15 T(4) = 4 * 15 + 4 = 64 … We can note a couple of things. First, the function … Read more
nested[1] and nested[2] have no effect whatsoever on the time taken to perform operations on nested[0]. An object has no knowledge of what containers it might be referenced in or what other objects might be in the container with it. List operations take the same amount of time regardless of what objects are being added, … Read more
Assuming that T(n) doesn’t suddenly become negative at some value of n, we can give a lower bound for the left hand side if we neglect the first term: We define a new function S(n) such that: We can immediately see that it has terms (ignoring off-by-one etc.). Thus if we keep expanding: At this … Read more