performance,time,time-complexity

You don't give a context what myarray, x and y are. Without that context, the question cannot be answered in any meaningful way. The extra assignments might have no side effects that cannot be optimised away. Basically, looking at speed optimisation at this elementary level is completely pointless. If you...

This will be not a very rigours analysis, but the problem seems to be with the BasicTransformer's transform(Seq[Node]) method[1]. The child transform method will be called twice for a node which is changed. Specifically in your example all the nodes will be called twice for this reason. And you are...

c#,optimization,time-complexity

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...

java,algorithm,time-complexity

Complexity is actually O(n^3), if ignoring JIT optimizations (and most possibly online judges turns that off). Since 5000^3 ~= 1.2*10^11, getting TLE is expected. Explanation for time complexity: Look at your code, and pay special attention to the comment I added: for(int i=0;i<n;i++){ String temp=""; for(int j=i;j<n;j++){ temp+=S.charAt(j); // ^^...

algorithm,iteration,time-complexity,recurrence,upperbound

Look at step 4 T(n) = 2(2(2(2(2T(n-5) + 1) + 1) + 1) + 1) +1 ----- 4 T(n) = 2(2(2(2(2T(n-5))))) + 16 + 8 + 4 + 2 +1 = T(n) = 2^4* 2T(n-5) + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 = T(n) = 2^4* 2T(n-5)...

java,algorithm,time-complexity,tail-recursion

The matrix has very simple structure (I draw only top right half, bottom left is the same, mirrored) 0 1 2 3 4 5 ... . 0 1 2 3 4 ... . . 0 1 2 3 ... . . . 0 1 2 ... . . . ....

The solution is indeed O(V^2): for (int i = 0; i < V; i++) dist[i] = INT_MAX, sptSet[i] = false; This part runs BEFORE the main loop, and in complexity of O(V) -. for (int count = 0; count < V-1; count++) { This is the main loop, it runs...

j is not reset to 0 with every iteration of the outer loop. As such, it runs to n-1 just once, same as i does. So you have two parallel/intermingled iterations from 0 to (at most) n-1. In every step, the program increases i by one. The program terminates when...

python,list,dictionary,set,time-complexity

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...

algorithm,loops,time-complexity,complexity-theory,probability

Let the length of the string be written as n. n = str.length(); Now, the outer for-loop iterates for the whole length of the String,i.e., n times. Hence,outer for-loop complexity is O(n). Talking about child loop, the inner while loop executes (val)^(1/62) times. So, you can consider the inner while-loop...

python,algorithm,big-o,time-complexity

The solve function turns one element after an other into #, in worst case until the whole grid contains only #. But since you start at a specific point in the grid and only allow the next # to be a direct neighbor, you do not get all the (N²)!...

algorithm,time-complexity,proof,dht,kademlia

It has been a while since I've actually read the paper, so I'm mostly piecing this together from my implementation experience instead of trying to match the concepts that I have in my head to the formal definitions in the paper, so take the following with a little grain of...

algorithm,big-o,time-complexity

They are coming from strict analysis of the complexity function. One common example is for Matrix Multiplication, while the Naive solution is O(n^3) multiply operations, there are some faster solutions. One of the first improvements offered to use 7 (instead of 8) multiplications operations to multiply two 2X2 matrices. If...

javascript,algorithm,time-complexity

First of all JavaScript does not require that objects be implemented as hashes (average O(1) lookup) rather than using some other structure like a BTree that is O(log(n)). So nobody can guarantee the performance of object attribute lookup. But usually people use hashes. Which are O(1). But as the joke...

math,big-o,time-complexity,asymptotic-complexity

I don't believe that the ordering you've given here is correct. Here are a few things to think about: Notice that 2log4 n = 2(log2 n / log2 4) = 2(log2 n) / 2. Can you simplify this expression? How fast does the function eπ4096 grow as a function of...

time-complexity,complexity-theory

Formal definitions of O notations are: f(n) = O(g(n)) means that there exists some constant c and n0 that f(n) <= c*g(n) for n >= n0. f(n) = Ω(g(n)) means that there exists some constant c and n0 that f(n) >= c*g(n) for n >= n0. f(n) = Θ(g(n)) means...

Time complexity and actual running time are two very different things. Time complexity only has meaning when we are talking about variable input size. It tells how well algorithm scales for larger inputs. If we assume that your input is n (or 1000000000 in the first two cases), then all...

c,algorithm,math,time-complexity

Since this is a Project Euler problem, you are supposed to be able to do it in about a minute of computing time on a modern computer. They don't always stick to that, but it indicates that a running time of k*n^2 or k*n^2*log(n) is probably fine if the constant...

python,max,time-complexity,complexity-theory

Recursive method decreases the input size by one each call, so it's theoretically linear (since you're essentially doing a linear search for the maximum). The implementation of Python's lists is going to distort a timer-based analysis.

For the first one, A: You need to se how many iterations the inner loop makes in total. So how many iterations does the inner loop have? It will have first 1, then 2, then 3 all the way until N^2. So we get: For the next one, B: Again,...

algorithm,math,time-complexity,computer-science,recurrence-relation

Here are a few hints : define R(n) = T(n)/(n-1)! solve the recurrence for R(n) express T(n) as a function of R(n) ...

loops,for-loop,big-o,time-complexity,asymptotic-complexity

If c=1 in loop 1 and loop 2 then it will run infinite times right but it is given as logarithimic time why? Yes you are right, if c = 1 then we will get infinite loops for both case 1 and case 2, so the condition "c is...

time-complexity,recurrence-relation

T(1) = O(1) T(n) = T(n-1) + O(1) This is indeed the signature of a sequential search and it has O(n) time complexity. This is because T(n) = T(n-1) has O(n) complexity and the O(1) term drops out. The solution is trivial: Say we know T(n) = T(n-1), and we...

java,arrays,algorithm,time-complexity

You can easily do this in O(n * log n) by sorting: int[] input = //... int[] input = new int[]{2, 6, 7, 3, 9}; Integer[] indices = new Integer[input.length]; for (int i = 0; i < indices.length; i++) { indices[i] = i; } // find permutation of indices that...

From your closed form of your formula, the term 1 / (sqrt 5) ((1 - sqrt 5) / 2)^n has limit 0 as n grows to infinity (|(1 - sqrt 5) / 2| < 1). Therefore we can ignore this term. Also since in time complexity theory we don't care...

mysql,hash,time-complexity,unique-index

A hash structure can identify non-collisions in hash buckets in O(1) time -- theoretically faster then a b-tree. Hashes are not O(n), unless "n" is the number of bits in a single key (usually it refers to the number of records). Collisions are a problem, because you have to test...

algorithm,math,recursion,time-complexity

As all the commenters said, I need to use the Akra-Bazzi theorem. C(1) = 1 C(2) = 1 For N > 2 we need to first find 'p' from the following equation : (1/3)^p + (2/3)^p = 1. It is obvious that p = 1. Next we need to solve...

time-complexity,complexity-theory

Don't have enough reputation points, hence posting as answer. Perhaps your use of factor in two different senses is the source of the confusion. Time is but one factor out of many possible complexity factors, such as storage, bandwidth, etc. Exponential factors in the case of polynomial algorithms refer to...

algorithm,time-complexity,complexity-theory,asymptotic-complexity,big-theta

Outermost loop will run for ceil(log n) times. The middle loop is dependent on the value of i. So, it's behaviour will be : 1st iteration of outermost-loop - 1 2nd iteration of outermost-loop - 2 ..................................... ceil(log n) iteration of outermost-loop - ceil(log n) Innermost loop is independent of...

c++,algorithm,inheritance,time-complexity

The first is supposedly in O(M*logN) time, where M is the size of the list, and N = number of concrete derived classes of Base It's not though. unordered_map is a hashtable, lookup and insertion have constant complexity on average. So the first is still O(M). Just with more...

c++,dictionary,time-complexity

Most probably you are thinking that std::map is a hash table which would have a linear complexity (worst case). std::map on the other hand is SORTED which complicates things a lot. By having that requirement you are restricted to data structures that can support sorting (usually red/black trees), from which...

python,python-2.7,time-complexity,space-complexity

map may be microscopically faster in some cases (when you're NOT making a lambda for the purpose, but using the same function in map and a listcomp). List comprehensions may be faster in other cases and most (not all) pythonistas consider them more direct and clearer. more clearly explained here...

if-statement,for-loop,time-complexity,asymptotic-complexity

We can start by looking at the easy case, case 2. Clearly, each time we go through the loop in case 2, one of either 2 things happens: count is incremented (which takes O(1) [except not really, but we just say it does for computers that operate on fixed-length numbers...

python,algorithm,time-complexity,longest-substring

It's O(N) 'why to use DP of O(N2)' : You don't need to for this problem. Note, though, that you take advantage of the fact that your sequence tokens (letters) are finite - so you can set up a list to hold all the possible starting values (26) and...

arrays,algorithm,random,insert,time-complexity

First consider the inner loop. When do we expect to have our first success (find an open position) when there are i values already in the array? For this we use the geometric distribution: Pr(X = k) = (1-p)^{k-1} p Where p is the probability of success for an attempt....

algorithm,time-complexity,complexity-theory

EDIT: I'll add a bit of explanation to clear up your confusion about the quote in your question. Let's consider a fixed value of i and focus on the innermost two loops: for (int j = 1; j <= i*i; j++) for (int k = 1; k <= j*j; k++)...

c++,algorithm,vector,time-complexity,binary-search-tree

It can be done in O(n) time and O(logN) space by doing an in-order traversal and stopping when you reach the n/2th node, just carry a counter that tells you how many nodes have been already traversed - no need to actually populate any vector. If you can modify your...

In your first example you have only one for loop. The i value linearly increases until the condition i<n*2; is met. The execution time of your for loop is linearly dependent on the value of n so its time complexity is O(n) because your i value is directly proportional to...

java,algorithm,runtime,big-o,time-complexity

Using that approach to building a heap, yes, but there is an O(n) algorithm for converting an array to a heap. See http://en.wikipedia.org/wiki/Binary_heap#Building_a_heap for details. That said, an O(m) time, O(n) memory solution exists for this problem, implemented by e.g. Guava's Ordering.leastOf. One implementation is create a buffer, an array...

ruby-on-rails,ruby,algorithm,hash,time-complexity

Simple benchmark: require 'benchmark' iterations = 10_000 small = 10 big = 1_000_000 small_hash = {} big_hash = {} (1..small).each do |i| small_hash[i] = i end (1..big).each do |i| big_hash[i] = i end Benchmark.bmbm do |bm| bm.report('Small Hash') do iterations.times { small_hash.has_key?(1) } end bm.report('Big Hash') do iterations.times { big_hash.has_key?(1)...

algorithm,math,time-complexity

From what I understand, you are being asked "when should I use which", should I use an algorithm that takes a constant time of 1 second? or should I use an algorithm that takes f(n) microseconds. Note that 1 second = 10^6 microseconds, so you are solving: f(n) <= 1,000,000...

You have 2 loops: one running 1.5n times, and the other running 1n times. The time complexity for that is 2.5n, which is O(n).

As other answers mentioned, there is only one stack and recursive calls evaluated strictly in given order. However, for analysis purpose, you can visualize entire call sequence as a tree. ...

arrays,list,tuples,time-complexity,sml

The sml function has O(1) complexity, so feel free to use it! e.x. `fun isTrue (n,tup) = if Array.sub(tup,n) then true else false;` As far as the tuple is concerned, you can only use a specific number, not a variable in a tuple. e.x. fun isTrue (n,tup) = if #2...

python,performance,algorithm,time-complexity,primes

The Sieve of Eratosthenes looks like this: def sieve(n): primality_flags = [True]*(n+1) flags[0] = flags[1] = False primes = [] for i, flag in enumerate(primality_flags): if flag: primes.append(i) for j in xrange(2*i, n+1, i): flags[i] = False It processes each number once when the outer loop reaches it, and once...

algorithm,big-o,time-complexity,complexity-theory,asymptotic-complexity

Sometimes you can simplify calculations by calculating the amount of time per item in the result rather than solving recurrence relations. That trick applies here. Start by changing the code to this obviously equivalent form: private static IntTreeNode createBST(int[] array, int left, int right) { int middle = array[(left +...

algorithm,time-complexity,complexity-theory,asymptotic-complexity

Given, f(x) = xlogx+3logx^2 = xlogx+6logx // since log (a^b) = b log a As we know, f(x) = O(g(x)), if | f(x) | <= M. | g(x) |, where M is a positive real number. Therefore, for M>=7 and x varying in the real positive range, M . x...

python,algorithm,dictionary,big-o,time-complexity

for n in nums: if n > least_max: max_nums[least_max_key] = n least_max_key = min(max_nums, key=max_nums.get) # this is O(k) least_max = max_nums[least_max_key] You're doing an O(k) operation n times, so the complexity of your second function is O(n*k). Assuming you want the output in sorted order, this can be done...

java,performance,algorithm,data-structures,time-complexity

A very rough estimation follows. According to wikipedia, the fastest supercomputer at the time of writing has 33.86 petaflops. We have: 33.86 petaflops = 33.86 * 10^15 ~= 3 * 10^16 FLOPS The recursive algorithm requires about phi^n operations, where phi is the golden ratio, equal to ~1.618. We have:...

algorithm,big-o,time-complexity,complexity-theory,exponential

Well, your method is also concrete. You should proceed in the same direction. Currently, I also don't have a better option. 4^n = ((2^2)^n) = (2^2n) = (2^n) * (2^n) > 2^n for all values of n>0. As, (2^n) * (2^n) > O(2^n) . This is because (2^n) * (2^n)...

The formal definition of big-O notation is f(x) ∈ O(g(x)) as there exists c > 0 (e.g., c = 1) and x0 (e.g., x0 = 5) such that f(x) < cg(x) whenever x > x0. From your thinking, it seems you don't understand that if there is a constant that...

javascript,time-complexity,immutable.js

In Immutable js source code, the key places that implement equality are deepEqual.js and is.js. The equality performs a recursive comparison into the map and does the comparison exactly once for each node. Thus the complexity of the comparison is O(n)....

java,sorting,time-complexity,radix-sort

for(int z = 0; z<bucket.length; z++){ for (int k=0; k<bucket[z].size(); k++){ The crucial thing here is that the sum total of all the bucket sizes will equal n, so these loops combined only take O(n). The loop over j takes O(n), and the while loop on i will run...

Question Does the time complexity of the code is O(n + k)? Or would it be O(n*k)? Neither. The complexity is O(n + k). In the case where n <= k, this would equal O(k), but this is not necessarily the case. n <= k (original answer) If the...

algorithm,search,time,time-complexity,depth-first-search

No you are not wrong, of course if you find the destination in current node's neighbors (children in your wordings), you can terminate it. However, I will stick to "standard" implementation due to 2 reasons: (Just my personal concerns) Easier Implementation, Higher Readability, For example, you may want to do...

algorithm,math,big-o,time-complexity,complexity-theory

No, you haven't made any mistake. You are correct in thinking so and is your solution perfect! The outer-loop will run n-times and the inner loop runs for 5(constant) times. Therefore, the loops complexity will be O(5*n) = O(n) and the other statements are of constant time-complexity. Since, 5*n times...

In your example, if num is prime then it would take exactly num - 1 steps. This would mean that the algorithm's runtime is O(num) (where O stands for a pessimistic case). But in case of algorithm that operate on numbers things get a little bit more tricky (thanks for...

The upper bound given by the other answers is actually too high. This algorithm has a O(n) runtime, which is a tighter upper bound than O(n*logn). Proof: Let's count how many total iterations the inner loop will perform. The outer loop runs n times. The inner loop runs at least...

algorithm,big-o,time-complexity

At least for practical purposes, the Landau O(...) can be viewed as a function (hence the appeal of its notation). This function has properties for standard operations, for example: O(f(x)) + O(g(x)) = O(f(x) + g(x)) O(f(x)) * O(g(x)) = O(f(x) * g(x)) O(k*f(x)) = O(f(x)) for well defined functions...

loops,for-loop,time-complexity,nested-loops,asymptotic-complexity

This question is tricky - there is a difference between what the runtime of the code is and what the return value is. The first loop's runtime is indeed O(log n), not O(log log n). I've reprinted it here: p = 0; for (j=n; j > 1; j=j/2) ++p; On...

In java you cannot return multiple values separately. What you could do is, you could add all the values to be returned to a list and then return the reference to the list. Something like this : List<Double> doubleList = new ArrayList<>(); doubleList.add(max); // index 0 doubleList.add(maxelement); // index 1...

time-complexity,asymptotic-complexity

If we consider C_1 and C_2 such that C_1 < C_2, then we can say the following with certainty (n^C_2)*log(n) grows faster than (n^C_1) This is because (n^C_1) grows slower than (n^C_2) (obviously) also, for values of n larger than 2 (for log in base 2), log(n) grows faster than...

java,arraylist,hashmap,time-complexity,asymptotic-complexity

Firstly, an operation of complexity O(1) does not always take lesser time than an operation of complexity O(n). O(1) only means that the operation takes a constant time (which could be any value), regardless of the size of the input. O(n) means that the time required for the operation increases...

algorithm,time-complexity,complexity-theory

It's easy to see from calculus that if lim {n -> inf} a(n) / b(n) < inf then a(n) = O(b(n)) Also note that all the functions here go to infinity, so we can use L'Hôpital's rule. Finally, note that, asymptotically, Stirling's Approximation gives lim {n -> inf} n! /...

algorithm,time-complexity,convex-hull

Well you do have linear complexity because: For (i=1 ... n) grants an n factor to the complexity so until now O(n) In the nested while loop you have the condition (L size >= 2 && it will also check if you do make a counter-clockwise turn(that should be done...

algorithm,big-o,time-complexity,complexity-theory

The purpose of the Big-O notation is to find what is the dominant factor in the asymptotic behavior of a function as the value tends towards the infinity. As we walk through the function domain, some factors become more important than others. Imagine f(n) = n^3+n^2. As n goes to...

arrays,data-structures,time-complexity

Here's an implementation that still has O(n) runtime for insert and delete, but which gets lookups running in time O(log n). Have as your data structure a dynamically-allocated, sorted array with no slack space. To perform a lookup, just use binary search in time O(log n). To do an insertion,...

Well the important thing to understand is how you're counting the complexity. Let's say you have N entries in the first container and the std::set<int> contains M elements. With a container that allows fast search, the first part is O(1) then printing is O(M). However, if you were using something...

search,big-o,time-complexity,complexity-theory,binary-search

For binary search, the array should be arranged in ascending or descending order. In each step, the algorithm compares the search key value with the key value of the middle element of the array. If the keys match, then a matching element has been found and its index, or position,...

python,algorithm,time-complexity,permutation

I believe there is a proof that the runtime is indeed O(n*n!). (Inspired by an earlier SO question here: complexity of recursive string permutation function) We have the following recursion for the time spent, without printing: T(n) = n*T(n-1) + O(n^2) Now if U(n) = T(n)/n! then we must have...

algorithm,big-o,time-complexity

The function in the code denoted as mergeSort() takes O(n) time, it is looping constant number of times over the elements within range (low,high). The function denoted as partition(), which is the actual sorting function, takes `O(nlogn) time. It can actually be denoted as: T(n) = 2T(n/2) + C*n //for...

java,algorithm,data-structures,big-o,time-complexity

Instead of calling Arrays.fill(answer, minimumValue); whenever you encounter a "max counter" operation, which takes O(N), you should keep track of the last max value that was assigned due to "max counter" operation, and update the entire array just one time, after all the operations are processed. This would take O(N+M)....

python,time-complexity,space-complexity

Without further information on the problem, your actual task and your solving attempts an answer could merely be adequate...but I will try to at least give you some input. a = [random.randint(1,100) for i in xrange(1000000)] A statement like a = ... is normally considered to have O(1) in terms...

algorithm,math,big-o,time-complexity,complexity-theory

Using O(n^1/4) is perfectly fine for big O notation. Here are some examples of fractures in exponents from real life examples O(n) is also correct (because big O giving only upper bound), but it is not tight, so n^1/4 is in O(n), but not in Theta(n) n^1/4 is NOT...

java,collections,time-complexity

Some implementations of LinkedList can keep a count of the number of elements in their list as well as a pointer to the last element for quick tail insertion. This means that they are done O(1) as it is a quick data access. Similar reasoning can be followed for implementations...

algorithm,union,time-complexity,dijkstra

So the time complexity of the algorithm is equal to the time complexity of the lines (3-9)+O(E). Which is the time complexity of the union? No, it is not the complexity of the union, union can be done pretty efficiently if you are using hash table for example. Moreover,...

algorithm,big-o,time-complexity,lower-bound

Yes, that's correct. One way to see this is via an adversarial argument. Suppose that you have an algorithm that allegedly finds the maximum value in the array, but doesn't inspect every array element at least once. Suppose I run your algorithm on some array A1 that consists of nothing...

hash,hashtable,big-o,time-complexity

When talking about hashing, we usually measure the performance of a hash table by talking about the expected number of probes that we need to make when searching for an element in the table. In most hashing setups, we can prove that the expected number of probes is O(1). Usually,...

big-o,time-complexity,complexity-theory,binary-search

As always, there can be a difference of the complexity of an algorithm and its implementation. If you implement binary search in a recursive function, that takes a copy of a part of the input array, this array has to be generated in O(n), so you are right, that this...

algorithm,sorting,big-o,time-complexity,complexity-theory

The outer for-loop would run for n times. But, it also holds an inner for-loop which is dependent on the vale of j. I think that the first for-loop test will execute n times, and the nested for-loop will execute 1 + 2 + ... + n = n(n+1)/2 times....

algorithm,recursion,time-complexity,asymptotic-complexity

Number 1 I can't tell how you've derived your formula. Usually adding terms happens when there are multiple steps in an algorithm, such as precomputing data and then looking up values from the data. Instead, nested for loops implies multiplication. Also, the worst case is the best case for this...

ruby-on-rails,algorithm,time-complexity

The solution is to sort the array, and then iterate it. You now only need to check for candidates that are adjacent (arr[i],arr[i+1]), and not every pair of elements. This runs in O(NlogN). Note that this is a generalization of Element Distinctness Problem, so if you are interested in worst...

algorithm,sorting,time-complexity,sse,simd

It sounds as though a sorting network is the answer to the question that you asked, since the position of the comparators is not data dependent. Batcher's bitonic mergesort is O(n log2 n).

Let's first consider the best-case complexity of determining the correct order of the elements in two sorted sets (without worrying about how they are stored). If the last element in the first set is smaller than the first element in the second set, only a single comparison is needed to...

lowerKey() is a search in a balanced binary search tree, so it's obviously O(log n). You might want to read the source code, e.g. from here, to see how the tree is traversed. Similarly, each operation with a NavigableMap returned from subMap() also requires O(log n) because you will need...

It is O(n). Big O is meant to describe the complexity of the application and in this case it is linear so it is O(n).

java,search,collections,time-complexity

The thing which is often skipped when comparing ArrayList and LinkedList is cache and memory management optimisations. ArrayList is effectively just an array which means that it is stored in a continuous space in the memory. This allows the Operating System to use optimisations such as "when a byte in...

c++,algorithm,recursion,time-complexity

T(n) = T(n-1) + c T(1) = c T(2) = c+c=2c T(3) = 2c +c=3c .... T(n) = nc Therefore time complexity is O(n)...

Since the O() behavior is asymptotic, you sometimes cannot see the behavior for small values of N. For example, if I set numberOfSizes = 9 and discard the first 3 points for the polynomial fit, the slope is much closer to 3: randomMPolyfit = polyfit(log10(sizesArray(4:end)), log10(randomMAveragesArray(4:end)), 1); randomMSlope = randomMPolyfit(1)...

algorithm,data-structures,big-o,time-complexity

You're mixing green apples with red apples here. Take your example, for instance - a Red-Black Tree's insert and remove methods keep it balanced. So when you search for a value, worst case O(log(n)) is because there's no sorting of any kind done. The tree is already "sorted" according to...

algorithm,language-agnostic,time-complexity,complexity-theory,analysis

Answering for the first part :- for i = 1 to 526 for j = 1 to n^2(lgn)^3 for k = 1 to n x=x+1 This program will run 526 * n^2 * (lg n)^3 * n times = 526 * n^3 * (lg n)^3 times. So, x = x...

math,time-complexity,complexity-theory

This statement is sometimes true and sometimes false depending on the base of the logarithm. Examples: O(2lg n) = O(n), where lg is the binary logarithm. O(2log_4 n) = O(n1/2), where log_4 is the logarithm to base 4. O(2log_1.25992 n) = O(n3), because 1.25992 is the cube root of 2....

java,performance,algorithm,big-o,time-complexity

when using O() notation you ignore constants, which means that O(n) == (10^10*n). so while O(n^2)>O(n) is true asymptotically, its not necessarily true for smaller values of n. in your case imagine that maybe resizing the array behind the hashset could be more time consuming than iterating the input.

The source code to the RTL is available with a purchase of Delphi, so it's easy to tell what's actually going on. Under the hood, items are managed inside a dynamic array that grows as you add items to it, but in a more intelligent fashion than simply growing by...

c++,time-complexity,sparse-matrix

nnz : non-zero number of sparse matrix row_size : matrix row number column_size : matrix column number There are many ways, their space complexity : Compressed Sparse Row (CSR) : 2*nnz + row_size number of memory Compressed Sparse Column (CSC) : 2*nnz + column_size number of memory Coordinate Format (COO)...