The famous mathematician Gauss is said to have found a formula for that exact problem when he was in primary school. And as mentioned by @Henry in the comments it is: Source: Wikipedia (DE), Wikipedia (EN) As work is done for every entry, i.e., O(1) is required for each “item”. Hence, the problem is in … Read more
Square root time complexity means that the algorithm requires O(N^(1/2)) evaluations where the size of input is N. As an example for an algorithm which takes O(sqrt(n)) time, Grover’s algorithm is one which takes that much time. Grover’s algorithm is a quantum algorithm for searching an unsorted database of n entries in O(sqrt(n)) time. Let … Read more
This follows easily from a simple fact in Calculus: and we have the following inequality: Here we can conclude that S = 1 + 1/2 + … + 1/n is both Ω(log(n)) and O(log(n)), thus it is Ɵ(log(n)), the bound is actually tight.
O(n!) isn’t equivalent to O(n^n). It is asymptotically less than O(n^n). O(log(n!)) is equal to O(n log(n)). Here is one way to prove that: Note that by using the log rule log(mn) = log(m) + log(n) we can see that: log(n!) = log(n*(n-1)*…2*1) = log(n) + log(n-1) + … log(2) + log(1) Proof that O(log(n!)) … Read more
In-order, Pre-order, and Post-order traversals are Depth-First traversals. For a Graph, the complexity of a Depth First Traversal is O(n + m), where n is the number of nodes, and m is the number of edges. Since a Binary Tree is also a Graph, the same applies here. The complexity of each of these Depth-first … Read more
O(max(input)+n) The complexity just appears awkward to express because most sorting algorithms are data agnostic. Their time scales with the amount of data, not the data itself. FWIW, as pointed out here, this is not a reliable algorithm for sorting data.
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
Short explanation: If an algorithm is of Θ(g(n)), it means that the running time of the algorithm as n (input size) gets larger is proportional to g(n). If an algorithm is of O(g(n)), it means that the running time of the algorithm as n gets larger is at most proportional to g(n). Normally, even when … Read more