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 am trying to solve the popular interview question Find the k-th smallest number in an array of distinct integers. I read some solutions and found that a heap data structure suits this problem very well.
So, I have tried to implement a solution using the PriorityQueue class of the Collections framework, assuming that it is functionally identical to a heap.
Here is the code that I've tried:
public static int getKthMinimum(int[] input, int k){
PriorityQueue<Integer> heap = new PriorityQueue<Integer>();
// Total cost of building the heap - O(n) or O(n log n) ?
for(int num : input){
heap.add(num); // Add to heap - O(log n)
}
// Total cost of removing k smallest elements - O(k log n) < O(n) ?
while(k > 0){
heap.poll(); // Remove min - O(log n)
k--;
}
return heap.peek(); // Fetch root - O(1)
}
Based on the docs, the poll & add methods take O(log n) time and peek takes constant time.
What will be the time complexity of the while loop? (I think O(k log n)).
Should O(k log n) be considered higher than O(n) for the purpose of this question? Is there a threshold where it switches?
What will be the total time complexity of this code? Will it be O(n)?
If not already O(n), is there a way to solve it in O(n), while using the PriorityQueue class?
1. What will be the time complexity of the while loop? (I think O(k log n)).
O(k log n) is correct.
2. Should O(k log n) be considered higher than O(n) for the purpose of this question? Is there a threshold where it switches?
You cannot assume that. k can be anywhere from 0 to n−1, which means that k log n can be anywhere from 0 to n log n.
3. What will be the total time complexity of this code? Will it be O(n)?
O(n log n), because that's the cost of building the heap.
It's possible to build a heap in O(n) time, but your code doesn't do that; if it did, your total complexity would be O(n + k log n) or, equivalently, O(MAX(n, k log n)).
4. If not already O(n), is there a way to solve it in O(n), while using the PriorityQueue class?
No. There exist selection algorithms in worst-case O(n) time, but they're a bit complicated, and they don't use PriorityQueue.
The fastest PriorityQueue-based solutions would require O(MAX(n, n log MIN(k, n−k))) time. (The key is to keep only the k smallest elements in the heap while iterating — or the n−k largest elements and use a max-heap, if k is large enough for that to be worthwhile.)
It's only checking the for loop 1/3n times, so it's still technically linear I guess? However I don't really understand why it wouldn't be O(logn), because many times a code with O(logn) running time ends up checking around 1/3n. Does O(logn) always divide the options by 2 every time?
int a = 0;
for (int i = 0; i < n; i = i+3)
a = a+i;
your code has complexity O(n), O(n)/3 == a * O(n) == O(n)
With time-complexity analysis, constant factors do not matter. You could do 1,000,000 operations per loop, and it will still be O(n). Because the constant 1/3 doesn't matter, it's still O(n). If you have n at 1,000,000, then 1/3 of n would be much bigger than log n.
From the Wikipedia entry on Big-O notation:
Let k be a constant. Then:
O(kg) = O(g) if k is nonzero.
It is order of n O(n) and not O(logn).
It because the run time increases linearly with the increase in n
For more information take a look at this graph and hopefully you will understand why it is not logn
https://www.cs.auckland.ac.nz/software/AlgAnim/fig/log_graph.gif
The running Time is O(n) (in unit complexity measure).
Does anyone know the running time in big O notation for the arrays.sort java method? I need this for my science fair project.
From official docs
I've observed that there are primarily two approaches. So, it depends on what you are sorting and what overloaded method from sort family of methods you are calling.
Docs mention that for primitive types such as long, byte (Ex: static void sort(long[])):
The sorting algorithm is a tuned quicksort, adapted from Jon L.
Bentley and M. Douglas McIlroy's "Engineering a Sort Function",
Software-Practice and Experience, Vol. 23(11) P. 1249-1265 (November
1993). This algorithm offers n*log(n) performance on many data sets
that cause other quicksorts to degrade to quadratic performance.
For Object types: (Ex: void sort(Object list[]))
Guaranteed O(nlogn) performance
The sorting algorithm is a modified mergesort (in which the merge is
omitted if the highest element in the low sublist is less than the
lowest element in the high sublist). This algorithm offers guaranteed
n*log(n) performance.
Hope that helps!
Arrays.sort() uses Tim sort - O(N log N) for array of objects and QuickSort for arrays of primitives - again O(N log N).
Here is an awesome comparison of sorting algorithms: http://www.sorting-algorithms.com/
I have tested the time complexity of Arrays.sort() in the various datasets. In the worst case, it is O(n^2) time complexity.
Trying doing this question using Arrays.sort() and then using Collections.sort(). You will see the difference. Question link
when I used Arrays.sort() it took more than 2 sec.
and when I used collections.sort() it took 0.2sec
Collections.sort() uses modified mergesort and Arrays.sort() uses QuickSort(). Below code is the best way to sort in java if you are using Array. In case of list you can use Collections.sort() by default.
static void sort(long[] a) {
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
It depends on what you are sorting and on your Java version. Sort methods for arrays of different types have different time and space complexities. There are also improvements in the latter Java versions.
Arrays.sort(int[])
As well as Arrays.sort(long[]), Arrays.sort(float[]) and Arrays.sort(double[])
Time complexity
Time complexity of Arrays.sort(int[]) depends on the version of Java.
Prior to Java 14
A pretty ordinary quicksort was used with time complexity ranging from O(n) (when the array is already sorted and we are only checking that it is) to O(n2) for certain inputs that cause extremely uneven distribution of elements into parts with an average complexity of O(n log(n)). You can find a detailed analysis here.
From Java 14 to Java 19
In Java 14 the implementation was improved to guarantee the worst-case time complexity of O(n log(n)). The function was changed to resort to heapsort if recursion becomes too deep:
if ((bits += DELTA) > MAX_RECURSION_DEPTH) {
heapSort(a, low, high);
return;
}
which prevents the method from degrading to quadratic time complexity.
Glimpse into the future
There is an initiative to switch to radix sort for almost random big enough arrays thus reducing the time complexity to O(n) in the worst-case.
Space complexity
In all versions, the algorithm has space complexity ranging from O(1) (when the array is already sorted and we only to check that it is) to O(n) (when the array is highly structured (there is a small number of sorted subarrays inside the original array and we merge those subarrays)).
Here's where allocation happens in the worst case:
/*
* Merge runs of highly structured array.
*/
if (count > 1) {
int[] b; int offset = low;
if (sorter == null || (b = (int[]) sorter.b) == null) {
b = new int[size];
} else {
offset = sorter.offset;
}
mergeRuns(a, b, offset, 1, sorter != null, run, 0, count);
}
return true;
DualPivotQuicksort.java
While the question asks specifically about Arrays.sort(int[]) method I still decided to include answers for other data types since this is the first result when you look for Arrays.sort() time and space complexity in Google and it is not easy to find correct answers to this simple question in other places.
Arrays.sort(short[])
As well as Arrays.sort(char[]) and Arrays.sort(byte[])
Time and space complexity
Although the documentation says:
The sorting algorithm is a Dual-Pivot Quicksort by Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm offers O(n log(n)) performance on all data sets, and is typically faster than traditional (one-pivot) Quicksort implementations.
This is not true at least starting from Java 7. Actually, an in-place counting sort used for big enough arrays, which has linear time complexity and constant space complexity:
private static void countingSort(short[] a, int low, int high) {
int[] count = new int[NUM_SHORT_VALUES];
/*
* Compute a histogram with the number of each values.
*/
for (int i = high; i > low; ++count[a[--i] & 0xFFFF]);
/*
* Place values on their final positions.
*/
if (high - low > NUM_SHORT_VALUES) {
for (int i = MAX_SHORT_INDEX; --i > Short.MAX_VALUE; ) {
int value = i & 0xFFFF;
for (low = high - count[value]; high > low;
a[--high] = (short) value
);
}
} else {
for (int i = MAX_SHORT_INDEX; high > low; ) {
while (count[--i & 0xFFFF] == 0);
int value = i & 0xFFFF;
int c = count[value];
do {
a[--high] = (short) value;
} while (--c > 0);
}
}
}
Counting sort implementation
Arrays.sort(Object[])
Unlike other methods, this one is well-documented and the documentation here corresponds to reality.
Time complexity
O(n log(n))
Starting from Java 7
This implementation is a stable, adaptive, iterative mergesort that requires far fewer than n lg(n) comparisons when the input array is partially sorted, while offering the performance of a traditional mergesort when the input array is randomly ordered. If the input array is nearly sorted, the implementation requires approximately n comparisons.
https://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html#sort(java.lang.Object[])
Before Java 7
The sorting algorithm is a modified mergesort (in which the merge is omitted if the highest element in the low sublist is less than the lowest element in the high sublist). This algorithm offers guaranteed n*log(n) performance.
https://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html#sort(java.lang.Object[])
Space complexity
O(n)
Starting from Java 7
Temporary storage requirements vary from a small constant for nearly sorted input arrays to n/2 object references for randomly ordered input arrays.
https://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html#sort(java.lang.Object[])
Before Java 7
The algorithm used by java.util.Arrays.sort and (indirectly) by java.util.Collections.sort to sort object references is a "modified mergesort (in which the merge is omitted if the highest element in the low sublist is less than the lowest element in the high sublist)." It is a reasonably fast stable sort that guarantees O(n log n) performance and requires O(n) extra space.
https://bugs.openjdk.org/browse/JDK-6804124