asymptotic-complexity

A min-max heap can be useful to implement a double-ended priority queue because of its constant time find-min and find-max operations. We can also retrieve the minimum and maximum elements in the min-max heap in O(log 2 n) time. Sometimes, though, we may also want to delete any node in the min-max heap, and this can be done in O(log 2 n) , according to the paper which introduced min-max heaps : ... The structure can also be generalized to support the operation Find(k) (determine the kth smallest value in the structure) in constant time and the operation Delete(k) (delete the kth smallest value

I was trying to find the complexities of an algorithm via different approaches. Mathematically I came across one O(m+n) and another O(mn) approach. However I am unable to grasp or say visualize this. It's not like I look at them and get the "Ahh! That's what's going on" feeling! Can someone explain this using their own examples or any other tool? My recommendation for finding intuition is thought experiments as follows: First, realize that m and n are two different measurements of the input . They might be the lengths of two input streams, the lengths of sides of a matrix, or the counts of two

A list of n strings each of length n is sorted into lexicographic order using the merge sort algorithm. The worst case running time of this computation is? I got this question as a homework. I know merge sort sorts in O(nlogn) time. For lexicographic order for length in is it n times nlogn ? or n^2 ? amit Each comparison of the algorithm is O(n) [comparing two strings is O(n) worst case - you might detect which is "bigger" only on the last character], You have O(nlogn) comparisons in mergesort. Thus you get O(nlogn * n) = O(n^2 * logn) But according to the recurrence relation T(n) = 2T(n/2) +

I am new in the algorithm analysis domain. I read here in the Stack Overflow question " What is a plain English explanation of "Big O" notation? " that O(2n^2) and O(100 n^2) are the same as O(n^2) . I don't understand this, because if we take n = 4, the number of operations will be: O(2 n^2) = 32 operations O(100 n^2) = 1600 operations O(n^2) = 16 operations Can any one can explain why we are supposed to treat these different operation counts as equivalent? Why this is true can be derived directly from the formal definition . More specifically, f(x) = O(g(n)) if and only if |f(x)| <= M|g(x)|

When analysing parallel algorithms, we tend to focus Work(T1), Span(T∞) or time. What I'm confused about is that if I was given an algorithm to analyse, what key hints would I need to look for, for Work, span and time? Suppose this algorithm: How do I analyse the above algorithm to find the Work, Time and span? Origin: PRAM s have been introduced in early 70-ies last century, in a hope it may allow to jump ahead a performance in tackling computationally hard problems. Yet, the promises or better expectations were cooled down by principal limitations these computing-device architectures still

I came across this question in which it was required to calculate in-degree of each node of a graph from its adjacency list representation. for each u for each Adj[i] where i!=u if (i,u) ∈ E in-degree[u]+=1 Now according to me its time complexity should be O(|V||E|+|V|^2) but the solution I referred instead described it to be equal to O(|V||E|) . Please help and tell me which one is correct. Corneliu Rather than O(|V||E|), the complexity of computing indegrees is O(|E|). Let us consider the following pseudocode for computing indegrees of each node: for each u indegree[u] = 0; for each u for

Is 2^n = Θ(4^n)? I'm pretty sure that 2^n is not in Ω(4^n) thus not in Θ(4^n), but my university tutor says it is. This confused me a lot and I couldn't find a clear answer per Google. chiwangc 2^n is NOT big-theta (Θ) of 4^n , this is because 2^n is NOT big-omega (Ω) of 4^n . By definition, we have f(x) = Θ(g(x)) if and only if f(x) = O(g(x)) and f(x) = Ω(g(x)) . Claim 2^n is not Ω(4^n) Proof Suppose 2^n = Ω(4^n) , then by definition of big-omega there exists constants c > 0 and n0 such that: 2^n ≥ c * 4^n for all n ≥ n0 By rearranging the inequality, we have: (1/2)^n ≥ c for all n ≥ n0 But