I have gone through Google and Stack Overflow search, but nowhere I was able to find a clear and straightforward explanation for how to calculate time complexity.
What do I know already?
Say for code as simple as the one below:
char h = 'y'; // This will be executed 1 time
int abc = 0; // This will be executed 1 time
Say for a loop like the one below:
for (int i = 0; i < N; i++) {
Console.Write('Hello, World!!');
}
int i=0; This will be executed only once.
The time is actually calculated to i=0 and not the declaration.
i < N; This will be executed N+1 times
i++ This will be executed N times
So the number of operations required by this loop are {1+(N+1)+N} = 2N+2. (But this still may be wrong, as I am not confident about my understanding.)
OK, so these small basic calculations I think I know, but in most cases I have seen the time complexity as O(N), O(n^2), O(log n), O(n!), and many others.
How to find time complexity of an algorithm
You add up how many machine instructions it will execute as a function of the size of its input, and then simplify the expression to the largest (when N is very large) term and can include any simplifying constant factor.
For example, lets see how we simplify 2N + 2 machine instructions to describe this as just O(N).
Why do we remove the two 2s ?
We are interested in the performance of the algorithm as N becomes large.
Consider the two terms 2N and 2.
What is the relative influence of these two terms as N becomes large? Suppose N is a million.
Then the first term is 2 million and the second term is only 2.
For this reason, we drop all but the largest terms for large N.
So, now we have gone from 2N + 2 to 2N.
Traditionally, we are only interested in performance up to constant factors.
This means that we don't really care if there is some constant multiple of difference in performance when N is large. The unit of 2N is not well-defined in the first place anyway. So we can multiply or divide by a constant factor to get to the simplest expression.
So 2N becomes just N.
This is an excellent article: Time complexity of algorithm
The below answer is copied from above (in case the excellent link goes bust)
The most common metric for calculating time complexity is Big O notation. This removes all constant factors so that the running time can be estimated in relation to N as N approaches infinity. In general you can think of it like this:
statement;
Is constant. The running time of the statement will not change in relation to N.
for ( i = 0; i < N; i++ )
statement;
Is linear. The running time of the loop is directly proportional to N. When N doubles, so does the running time.
for ( i = 0; i < N; i++ ) {
for ( j = 0; j < N; j++ )
statement;
}
Is quadratic. The running time of the two loops is proportional to the square of N. When N doubles, the running time increases by N * N.
while ( low <= high ) {
mid = ( low + high ) / 2;
if ( target < list[mid] )
high = mid - 1;
else if ( target > list[mid] )
low = mid + 1;
else break;
}
Is logarithmic. The running time of the algorithm is proportional to the number of times N can be divided by 2. This is because the algorithm divides the working area in half with each iteration.
void quicksort (int list[], int left, int right)
{
int pivot = partition (list, left, right);
quicksort(list, left, pivot - 1);
quicksort(list, pivot + 1, right);
}
Is N * log (N). The running time consists of N loops (iterative or recursive) that are logarithmic, thus the algorithm is a combination of linear and logarithmic.
In general, doing something with every item in one dimension is linear, doing something with every item in two dimensions is quadratic, and dividing the working area in half is logarithmic. There are other Big O measures such as cubic, exponential, and square root, but they're not nearly as common. Big O notation is described as O ( <type> ) where <type> is the measure. The quicksort algorithm would be described as O (N * log(N )).
Note that none of this has taken into account best, average, and worst case measures. Each would have its own Big O notation. Also note that this is a VERY simplistic explanation. Big O is the most common, but it's also more complex that I've shown. There are also other notations such as big omega, little o, and big theta. You probably won't encounter them outside of an algorithm analysis course. ;)
Taken from here - Introduction to Time Complexity of an Algorithm
1. Introduction
In computer science, the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the string representing the input.
2. Big O notation
The time complexity of an algorithm is commonly expressed using big O notation, which excludes coefficients and lower order terms. When expressed this way, the time complexity is said to be described asymptotically, i.e., as the input size goes to infinity.
For example, if the time required by an algorithm on all inputs of size n is at most 5n3 + 3n, the asymptotic time complexity is O(n3). More on that later.
A few more examples:
1 = O(n)
n = O(n2)
log(n) = O(n)
2 n + 1 = O(n)
3. O(1) constant time:
An algorithm is said to run in constant time if it requires the same amount of time regardless of the input size.
Examples:
array: accessing any element
fixed-size stack: push and pop methods
fixed-size queue: enqueue and dequeue methods
4. O(n) linear time
An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i.e. time grows linearly as input size increases.
Consider the following examples. Below I am linearly searching for an element, and this has a time complexity of O(n).
int find = 66;
var numbers = new int[] { 33, 435, 36, 37, 43, 45, 66, 656, 2232 };
for (int i = 0; i < numbers.Length - 1; i++)
{
if(find == numbers[i])
{
return;
}
}
More Examples:
Array: Linear Search, Traversing, Find minimum etc
ArrayList: contains method
Queue: contains method
5. O(log n) logarithmic time:
An algorithm is said to run in logarithmic time if its time execution is proportional to the logarithm of the input size.
Example: Binary Search
Recall the "twenty questions" game - the task is to guess the value of a hidden number in an interval. Each time you make a guess, you are told whether your guess is too high or too low. Twenty questions game implies a strategy that uses your guess number to halve the interval size. This is an example of the general problem-solving method known as binary search.
6. O(n2) quadratic time
An algorithm is said to run in quadratic time if its time execution is proportional to the square of the input size.
Examples:
Bubble Sort
Selection Sort
Insertion Sort
7. Some useful links
Big-O Misconceptions
Determining The Complexity Of Algorithm
Big O Cheat Sheet
Several examples of loop.
O(n) time complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount. For example following functions have O(n) time complexity.
// Here c is a positive integer constant
for (int i = 1; i <= n; i += c) {
// some O(1) expressions
}
for (int i = n; i > 0; i -= c) {
// some O(1) expressions
}
O(nc) time complexity of nested loops is equal to the number of times the innermost statement is executed. For example, the following sample loops have O(n2) time complexity
for (int i = 1; i <=n; i += c) {
for (int j = 1; j <=n; j += c) {
// some O(1) expressions
}
}
for (int i = n; i > 0; i += c) {
for (int j = i+1; j <=n; j += c) {
// some O(1) expressions
}
For example, selection sort and insertion sort have O(n2) time complexity.
O(log n) time complexity of a loop is considered as O(log n) if the loop variables is divided / multiplied by a constant amount.
for (int i = 1; i <=n; i *= c) {
// some O(1) expressions
}
for (int i = n; i > 0; i /= c) {
// some O(1) expressions
}
For example, [binary search][3] has _O(log n)_ time complexity.
O(log log n) time complexity of a loop is considered as O(log log n) if the loop variables is reduced / increased exponentially by a constant amount.
// Here c is a constant greater than 1
for (int i = 2; i <=n; i = pow(i, c)) {
// some O(1) expressions
}
//Here fun is sqrt or cuberoot or any other constant root
for (int i = n; i > 0; i = fun(i)) {
// some O(1) expressions
}
One example of time complexity analysis
int fun(int n)
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j < n; j += i)
{
// Some O(1) task
}
}
}
Analysis:
For i = 1, the inner loop is executed n times.
For i = 2, the inner loop is executed approximately n/2 times.
For i = 3, the inner loop is executed approximately n/3 times.
For i = 4, the inner loop is executed approximately n/4 times.
…………………………………………………….
For i = n, the inner loop is executed approximately n/n times.
So the total time complexity of the above algorithm is (n + n/2 + n/3 + … + n/n), which becomes n * (1/1 + 1/2 + 1/3 + … + 1/n)
The important thing about series (1/1 + 1/2 + 1/3 + … + 1/n) is around to O(log n). So the time complexity of the above code is O(n·log n).
References:
1
2
3
Time complexity with examples
1 - Basic operations (arithmetic, comparisons, accessing array’s elements, assignment): The running time is always constant O(1)
Example:
read(x) // O(1)
a = 10; // O(1)
a = 1,000,000,000,000,000,000 // O(1)
2 - If then else statement: Only taking the maximum running time from two or more possible statements.
Example:
age = read(x) // (1+1) = 2
if age < 17 then begin // 1
status = "Not allowed!"; // 1
end else begin
status = "Welcome! Please come in"; // 1
visitors = visitors + 1; // 1+1 = 2
end;
So, the complexity of the above pseudo code is T(n) = 2 + 1 + max(1, 1+2) = 6. Thus, its big oh is still constant T(n) = O(1).
3 - Looping (for, while, repeat): Running time for this statement is the number of loops multiplied by the number of operations inside that looping.
Example:
total = 0; // 1
for i = 1 to n do begin // (1+1)*n = 2n
total = total + i; // (1+1)*n = 2n
end;
writeln(total); // 1
So, its complexity is T(n) = 1+4n+1 = 4n + 2. Thus, T(n) = O(n).
4 - Nested loop (looping inside looping): Since there is at least one looping inside the main looping, running time of this statement used O(n^2) or O(n^3).
Example:
for i = 1 to n do begin // (1+1)*n = 2n
for j = 1 to n do begin // (1+1)n*n = 2n^2
x = x + 1; // (1+1)n*n = 2n^2
print(x); // (n*n) = n^2
end;
end;
Common running time
There are some common running times when analyzing an algorithm:
O(1) – Constant time
Constant time means the running time is constant, it’s not affected by the input size.
O(n) – Linear time
When an algorithm accepts n input size, it would perform n operations as well.
O(log n) – Logarithmic time
Algorithm that has running time O(log n) is slight faster than O(n). Commonly, algorithm divides the problem into sub problems with the same size. Example: binary search algorithm, binary conversion algorithm.
O(n log n) – Linearithmic time
This running time is often found in "divide & conquer algorithms" which divide the problem into sub problems recursively and then merge them in n time. Example: Merge Sort algorithm.
O(n2) – Quadratic time
Look Bubble Sort algorithm!
O(n3) – Cubic time
It has the same principle with O(n2).
O(2n) – Exponential time
It is very slow as input get larger, if n = 1,000,000, T(n) would be 21,000,000. Brute Force algorithm has this running time.
O(n!) – Factorial time
The slowest!!! Example: Travelling salesman problem (TSP)
It is taken from this article. It is very well explained and you should give it a read.
When you're analyzing code, you have to analyse it line by line, counting every operation/recognizing time complexity. In the end, you have to sum it to get whole picture.
For example, you can have one simple loop with linear complexity, but later in that same program you can have a triple loop that has cubic complexity, so your program will have cubic complexity. Function order of growth comes into play right here.
Let's look at what are possibilities for time complexity of an algorithm, you can see order of growth I mentioned above:
Constant time has an order of growth 1, for example: a = b + c.
Logarithmic time has an order of growth log N. It usually occurs when you're dividing something in half (binary search, trees, and even loops), or multiplying something in same way.
Linear. The order of growth is N, for example
int p = 0;
for (int i = 1; i < N; i++)
p = p + 2;
Linearithmic. The order of growth is n·log N. It usually occurs in divide-and-conquer algorithms.
Cubic. The order of growth is N3. A classic example is a triple loop where you check all triplets:
int x = 0;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
for (int k = 0; k < N; k++)
x = x + 2
Exponential. The order of growth is 2N. It usually occurs when you do exhaustive search, for example, check subsets of some set.
Loosely speaking, time complexity is a way of summarising how the number of operations or run-time of an algorithm grows as the input size increases.
Like most things in life, a cocktail party can help us understand.
O(N)
When you arrive at the party, you have to shake everyone's hand (do an operation on every item). As the number of attendees N increases, the time/work it will take you to shake everyone's hand increases as O(N).
Why O(N) and not cN?
There's variation in the amount of time it takes to shake hands with people. You could average this out and capture it in a constant c. But the fundamental operation here --- shaking hands with everyone --- would always be proportional to O(N), no matter what c was. When debating whether we should go to a cocktail party, we're often more interested in the fact that we'll have to meet everyone than in the minute details of what those meetings look like.
O(N^2)
The host of the cocktail party wants you to play a silly game where everyone meets everyone else. Therefore, you must meet N-1 other people and, because the next person has already met you, they must meet N-2 people, and so on. The sum of this series is x^2/2+x/2. As the number of attendees grows, the x^2 term gets big fast, so we just drop everything else.
O(N^3)
You have to meet everyone else and, during each meeting, you must talk about everyone else in the room.
O(1)
The host wants to announce something. They ding a wineglass and speak loudly. Everyone hears them. It turns out it doesn't matter how many attendees there are, this operation always takes the same amount of time.
O(log N)
The host has laid everyone out at the table in alphabetical order. Where is Dan? You reason that he must be somewhere between Adam and Mandy (certainly not between Mandy and Zach!). Given that, is he between George and Mandy? No. He must be between Adam and Fred, and between Cindy and Fred. And so on... we can efficiently locate Dan by looking at half the set and then half of that set. Ultimately, we look at O(log_2 N) individuals.
O(N log N)
You could find where to sit down at the table using the algorithm above. If a large number of people came to the table, one at a time, and all did this, that would take O(N log N) time. This turns out to be how long it takes to sort any collection of items when they must be compared.
Best/Worst Case
You arrive at the party and need to find Inigo - how long will it take? It depends on when you arrive. If everyone is milling around you've hit the worst-case: it will take O(N) time. However, if everyone is sitting down at the table, it will take only O(log N) time. Or maybe you can leverage the host's wineglass-shouting power and it will take only O(1) time.
Assuming the host is unavailable, we can say that the Inigo-finding algorithm has a lower-bound of O(log N) and an upper-bound of O(N), depending on the state of the party when you arrive.
Space & Communication
The same ideas can be applied to understanding how algorithms use space or communication.
Knuth has written a nice paper about the former entitled "The Complexity of Songs".
Theorem 2: There exist arbitrarily long songs of complexity O(1).
PROOF: (due to Casey and the Sunshine Band). Consider the songs Sk defined by (15), but with
V_k = 'That's the way,' U 'I like it, ' U
U = 'uh huh,' 'uh huh'
for all k.
For the mathematically-minded people: The master theorem is another useful thing to know when studying complexity.
O(n) is big O notation used for writing time complexity of an algorithm. When you add up the number of executions in an algorithm, you'll get an expression in result like 2N+2. In this expression, N is the dominating term (the term having largest effect on expression if its value increases or decreases). Now O(N) is the time complexity while N is dominating term.
Example
For i = 1 to n;
j = 0;
while(j <= n);
j = j + 1;
Here the total number of executions for the inner loop are n+1 and the total number of executions for the outer loop are n(n+1)/2, so the total number of executions for the whole algorithm are n + 1 + n(n+1/2) = (n2 + 3n)/2.
Here n^2 is the dominating term so the time complexity for this algorithm is O(n2).
Other answers concentrate on the big-O-notation and practical examples. I want to answer the question by emphasizing the theoretical view. The explanation below is necessarily lacking in details; an excellent source to learn computational complexity theory is Introduction to the Theory of Computation by Michael Sipser.
Turing Machines
The most widespread model to investigate any question about computation is a Turing machine. A Turing machine has a one dimensional tape consisting of symbols which is used as a memory device. It has a tapehead which is used to write and read from the tape. It has a transition table determining the machine's behaviour, which is a fixed hardware component that is decided when the machine is created. A Turing machine works at discrete time steps doing the following:
It reads the symbol under the tapehead.
Depending on the symbol and its internal state, which can only take finitely many values, it reads three values s, σ, and X from its transition table, where s is an internal state, σ is a symbol, and X is either Right or Left.
It changes its internal state to s.
It changes the symbol it has read to σ.
It moves the tapehead one step according to the direction in X.
Turing machines are powerful models of computation. They can do everything that your digital computer can do. They were introduced before the advent of digital modern computers by the father of theoretical computer science and mathematician: Alan Turing.
Time Complexity
It is hard to define the time complexity of a single problem like "Does white have a winning strategy in chess?" because there is a machine which runs for a single step giving the correct answer: Either the machine which says directly 'No' or directly 'Yes'. To make it work we instead define the time complexity of a family of problems L each of which has a size, usually the length of the problem description. Then we take a Turing machine M which correctly solves every problem in that family. When M is given a problem of this family of size n, it solves it in finitely many steps. Let us call f(n) the longest possible time it takes M to solve problems of size n. Then we say that the time complexity of L is O(f(n)), which means that there is a Turing machine which will solve an instance of it of size n in at most C.f(n) time where C is a constant independent of n.
Isn't it dependent on the machines? Can digital computers do it faster?
Yes! Some problems can be solved faster by other models of computation, for example two tape Turing machines solve some problems faster than those with a single tape. This is why theoreticians prefer to use robust complexity classes such as NL, P, NP, PSPACE, EXPTIME, etc. For example, P is the class of decision problems whose time complexity is O(p(n)) where p is a polynomial. The class P do not change even if you add ten thousand tapes to your Turing machine, or use other types of theoretical models such as random access machines.
A Difference in Theory and Practice
It is usually assumed that the time complexity of integer addition is O(1). This assumption makes sense in practice because computers use a fixed number of bits to store numbers for many applications. There is no reason to assume such a thing in theory, so time complexity of addition is O(k) where k is the number of bits needed to express the integer.
Finding The Time Complexity of a Class of Problems
The straightforward way to show the time complexity of a problem is O(f(n)) is to construct a Turing machine which solves it in O(f(n)) time. Creating Turing machines for complex problems is not trivial; one needs some familiarity with them. A transition table for a Turing machine is rarely given, and it is described in high level. It becomes easier to see how long it will take a machine to halt as one gets themselves familiar with them.
Showing that a problem is not O(f(n)) time complexity is another story... Even though there are some results like the time hierarchy theorem, there are many open problems here. For example whether problems in NP are in P, i.e. solvable in polynomial time, is one of the seven millennium prize problems in mathematics, whose solver will be awarded 1 million dollars.
I found some very helpful solutions to writing a Javascript function that recursively checks whether a string is a palindrome here. I would like to know what the time and space complexity of the following solution would be. Can you explain how each line contributes to big O?
function isPalindrome(string) {
if (string.length < 2) return true;
if (string[0] === string[string.length - 1]) {
return isPalindrome(string.slice(1, string.length - 1))
}
return false;
}
At first it would seem that your function is O(n), because you call it recursively n/2 times. However, in each call you also use string.slice, which has a complexity of O(n). Because of that, your function is actually O(n^2)
You recursively call the function n/2 times, where n is the length of the string, since you remove first and last entry of the string at each iteration.
Therefore, the complexity would be O(n/2) = O(n), and you have at most 3 operations at each function call. That would multiply all this by a constant, which does not change the complexity.
EDIT: as noted in comments and another answer, one of these operations is string.slice. You need to check for the complexity of this, because it will multiply as well and can change overall complexity. If slice is O(1) constant, then overall you'll have O(n). If slice is O(n), then overall that would be O(n^2).
For space complexity, you create a lot of arrays. I'll let you the details, but I'm tempted to say it's O(n^2) at first sight.
Sketch of calculations : first array is size n, then n-2, n-4, sum that all up with some math summation formulas.
hint : n + (n-1) + (n-2) + ... is n * (n + 1) / 2 which is O(n^2), that should give enough "feel" that this is O(n^2) too.
Update:
- Wikipedia says o(n) not O(n) so this algorithm isn't an in-place solution.
- Debating whether this solution runs in O(n) or O(1).
This is my solution for LeetCode question Flatten Binary Tree to Linked List. In terms of time and space complexity analysis, I'm not quite sure it's O(1) according to some explanations. I think it's O(n) because of the use of the stack. Is my analysis correct?
According to Wikipedia, O(n) is still accepted as in-place.
More broadly, in-place means that the algorithm does not use extra space for manipulating the input but may require a small though nonconstant extra space for its operation. Usually, this space is O(log n), though sometimes anything in o(n) is allowed.
/**
* Function flattens a binary tree.
* Time = O(n) because we iterate through all nodes.
* Space = O(n) because we use a stack.
* #param {root} root Input tree.
*/
var flatten = function(root) {
// If reach end of leaf node, return.
if (root === null) return;
let stack = [];
let currentNode = root;
while (true) {
// Push right branch to stack first,
if (currentNode.right) stack.push(currentNode.right);
// then left branch.
if (currentNode.left) stack.push(currentNode.left);
// If there are branches in stack:
if (stack.length > 0) {
// Change the current currentNode right branch to the last element of the stack:
currentNode.right = stack.pop();
// Change left branch to null
currentNode.left = null;
// Advance by changing current currentNode to the right currentNode.
currentNode = currentNode.right;
}
else break;
}
}
Yes, the algorithm uses Theta(N) worst-case time and worst-case additional space where N is the number of nodes in the tree. The time complexity is is clear, because you push and pop each node once.
The space complexity is a bit more tricky, but consider for example a tree that is a list where the next elements of the list are the left children, but assume it has additionally for each of these list nodes one right child which itself does not have any children.
With that example your algorithm will walk through the left-hand nodes, adding the right-hand ones to the stack, but only popping those once the end of the original list is reached, i.e. there will be about N/2 elements in the stack.
Best-case, the time complexity is still Theta(N), because you always iterate over all nodes, but the best-case space complexity is Theta(1), because e.g. a tree that is already a list will never increase the stack size to more than 1.
This would usually not be considered an in-place algorithm, because it uses additional Theta(N) space, at least worst-case. As explained in the Wikipedia article, an in-place algorithm should require o(N) additional space or, I would say, usually only O(1) or slightly more than that, e.g. O((ln N)^k) for some k maximum. Whether one should count worst-case or average-case is another question.
o(N) is the little-o notation, not the big-O notation. It means the the time/space is asymptotically smaller than c*N for every c > 0. Theta(N) is therefore never o(N).
Also, Theta(N) means that not only is it O(N), meaning asymptotically smaller then c*N for some c > 0, but also that it is asymptotically larger than c*N for some c.
If you want an in-place algorithm by the stricter definition, you should not require any additional container of dynamic size.
I want to create an array of type number with length n. All values inside the array should be 0 except the one which index matches a condition.
Thats how i currently do it:
const data: number[] = [];
for (let i = 0; i < n; i++) {
if (i === someIndex) {
data.push(someNumber);
} else {
data.push(0);
}
}
So lets say n = 4, someIndex = 2, someNumber = 4 would result in the array [0, 0, 4, 0].
Is there a way to do it in O(1) instead of O(n)?
Creating an array of size n in O(1) time is theoretically possible depending on implementation details - in principle, if an array is implemented as a hashtable then its length property can be set without allocating or initialising space for all of its elements. The ECMAScript specification for the Array(n) constructor doesn't mandate that Array(n) should do anything which necessarily takes more than O(1) time, although it also doesn't mandate that the time complexity is O(1).
In practice, Array(n)'s time complexity depends on the browser, though verifying this is a bit tricky. The performance.now() function can be used to measure the time elapsed between the start and end of a computation, but the precision of this function is artificially reduced in many browsers to protect against CPU-timing attacks like Spectre. To get around this, we can call the constructor repetitions times, and then divide the time elapsed by repetitions to get a more precise measurement per constructor call.
My timing code is below:
function timeArray(n, repetitions=100000) {
var startTime = performance.now();
for(var i = 0; i < repetitions; ++i) {
var arr = Array(n);
arr[n-1] = 'foo';
}
var endTime = performance.now();
return (endTime - startTime) / repetitions;
}
for(var n = 10000; n <= 1000000; n += 10000) {
console.log(n, timeArray(n));
}
Here's my results from Google Chrome (version 74) and Firefox (version 72); on Chrome the performance is clearly O(n) and on Firefox it's clearly O(1) with a quite consistent time of about 0.01ms on my machine.
I measured using repetitions = 1000 on Chrome, and repetitions = 100000 on Firefox, to get accurate enough results within a reasonable time.
Another option proposed by #M.Dietz in the comments is to declare the array like var arr = []; and then assign at some index (e.g. arr[n-1] = 'foo';). This turns out to take O(1) time on both Chrome and Firefox, both consistently under one nanosecond:
That suggests the version using [] is better to use than the version using Array(n), but still the specification doesn't mandate that this should take O(1) time, so there may be other browsers where this version takes O(n) time. If anybody gets different results on another browser (or another version of one of these browsers) then please do add a comment.
You need to assign n values, and so there is that amount of work to do. The work increases linearly with increasing n.
Having said that, you can hope to make your code a bit faster by making use of .fill:
const data: number[] = Array(n).fill(0);
data[someIndex] = someNumber;
But don't be mistaken; this is still O(n): .fill may be faster, but it still requires to fill the whole array with zeroes, which means a corresponding size of memory needs to be initialised, so that operation has linear time complexity.
If however you drop the requirement that zeroes need to be assigned, then you can only store the someNumber:
const data: number[] = Array(n);
data[someIndex] = someNumber;
This way you actually do not allocate the memory for the whole array, so this code snippet runs in constant time. Any access to an index different from someIndex will give you a value of undefined. You may trap that condition and translate that to a zero on-the-fly:
let value = i in data ? data[i] : 0;
Obviously, if you are going to access all indices of the array like that, you'll have again a linear time complexity.
If I remove one element from an array using splice() like so:
arr.splice(i, 1);
Will this be O(n) in the worst case because it shifts all the elements after i? Or is it constant time, with some linked list magic underneath?
Worst case should be O(n) (copying all n-1 elements to new array).
A linked list would be O(1) for a single deletion.
For those interested I've made this lazily-crafted benchmark. (Please don't run on Windows XP/Vista). As you can see from this though, it looks fairly constant (i.e. O(1)), so who knows what they're doing behind the scenes to make this crazy-fast. Note that regardless, the actual splice is VERY fast.
Rerunning an extended benchmark directly in the V8 shell that suggest O(n). Note though that you need huge array sizes to get a runtime that's likely to affect your code. This should be expected as if you look at the V8 code it uses memmove to create the new array.
The Test:
I took the advice in the comments and wrote a simple test to time splicing a data-set array of size 3,000, each one containing 3,000 items in it. The test would simply splice the
first item in the first array
second item in the second array
third item in the third array
...
3000th item in the 3000th array
I pre-built the array to keep things simple.
The Findings:
The weirdest thing is that the number of times where the process of the splice even takes longer than 1ms grows linearly as you increase the size of the dataset.
I went as far as testing it for a dataset of 300,000 on my machine (but the SO snippet tends to crash after 3,000).
I also noticed that the number of splice()s that took longer than 1ms for a given dataset (30,000 in my case) was random. So I ran the test 1,000 times and plotted the number of results, and it looked like a standard distribution; leading me to believe that the randomness was just caused by the scheduler interrupts.
This goes against my hypothesis and #Ivan's guess that splice()ing from the beginning of an array will have a O(n) time complexity
Below is my test:
let data = []
const results = []
const dataSet = 3000
function spliceIt(i) {
data[i].splice(i, 1)
}
function test() {
for (let i=0; i < dataSet; i++) {
let start = Date.now()
spliceIt(i);
let end = Date.now()
results.push(end - start)
}
}
function setup() {
data = (new Array(dataSet)).fill().map(arr => new Array(dataSet).fill().map(el => 0))
}
setup()
test()
// console.log("data before test", data)
// console.log("data after test", data)
// console.log("all results: ", results)
console.log("results that took more than 1ms: ", results.filter(r => r >= 1))
¡Hi!
I did an experiment myself and would like to share my findings. The experiment was very simple, we ran 100 splice operations on an array of size n, and calculate the average time each splice function took. Then we varied the size of n, to check how it behave.
This graph summarizes our findings for big numbers:
For big numbers it seems to behave linearly.
We also checked with "small" numbers (they were still quite big but not as big):
On this case it seems to be constant.
If I would have to decide for one option I would say it is O(n), because that is how it behaves for big numbers. Bear in mind though, that the linear behaviour only shows for VERY big numbers.
However, It is hard to go for a definitive answer because the array implementation in javascript dependes A LOT on how the array is declared and manipulated.
I recommend this stackoverflow discussion and this quora discussion to understand how arrays work.
I run it in node v10.15.3 and the code used is the following:
const f = async () => {
const n = 80000000;
const tries = 100;
const array = [];
for (let i = 0; i < n; i++) { // build initial array
array.push(i);
}
let sum = 0;
for (let i = 0; i < tries; i++) {
const index = Math.floor(Math.random() * (n));
const start = new Date();
array.splice(index, 1); // UNCOMMENT FOR OPTION A
// array.splice(index, 0, -1); // UNCOMMENT FOR OPTION B
const time = new Date().getTime() - start.getTime();
sum += time;
array.push(-2); // UNCOMMENT FOR OPTION A, to keep it of size n
// array.pop(); // UNCOMMENT FOR OPTION B, to keep it of size n
}
console.log('for an array of size', n, 'the average time of', tries, 'splices was:', sum / tries);
};
f();
Note that the code has an Option B, we did the same experiment for the three argument splice function to insert an element. It worked similary.