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Let's say that the algorithm involves iterating through a string character by character.
If I know for sure that the length of the string is less than, say, 15 characters, will the time complexity be O(1) or will it remain as O(n)?
There are two aspects to this question - the core of the question is, can problem constraints change the asymptotic complexity of an algorithm? The answer to that is yes. But then you give an example of a constraint (strings limited to 15 characters) where the answer is: the question doesn't make sense. A lot of the other answers here are misleading because they address only the second aspect but try to reach a conclusion about the first one.
Formally, the asymptotic complexity of an algorithm is measured by considering a set of inputs where the input sizes (i.e. what we call n) are unbounded. The reason n must be unbounded is because the definition of asymptotic complexity is a statement like "there is some n0 such that for all n ≥ n0, ...", so if the set doesn't contain any inputs of size n ≥ n0 then this statement is vacuous.
Since algorithms can have different running times depending on which inputs of each size we consider, we often distinguish between "average", "worst case" and "best case" time complexity. Take for example insertion sort:
In the average case, insertion sort has to compare the current element with half of the elements in the sorted portion of the array, so the algorithm does about n2/4 comparisons.
In the worst case, when the array is in descending order, insertion sort has to compare the current element with every element in the sorted portion (because it's less than all of them), so the algorithm does about n2/2 comparisons.
In the best case, when the array is in ascending order, insertion sort only has to compare the current element with the largest element in the sorted portion, so the algorithm does about n comparisons.
However, now suppose we add the constraint that the input array is always in ascending order except for its smallest element:
Now the average case does about 3n/2 comparisons,
The worst case does about 2n comparisons,
And the best case does about n comparisons.
Note that it's the same algorithm, insertion sort, but because we're considering a different set of inputs where the algorithm has different performance characteristics, we end up with a different time complexity for the average case because we're taking an average over a different set, and similarly we get a different time complexity for the worst case because we're choosing the worst inputs from a different set. Hence, yes, adding a problem constraint can change the time complexity even if the algorithm itself is not changed.
However, now let's consider your example of an algorithm which iterates over each character in a string, with the added constraint that the string's length is at most 15 characters. Here, it does not make sense to talk about the asymptotic complexity, because the input sizes n in your set are not unbounded. This particular set of inputs is not valid for doing such an analysis with.
In the mathematical sense, yes. Big-O notation describes the behavior of an algorithm in the limit, and if you have a fixed upper bound on the input size, that implies it has a maximum constant complexity.
That said, context is important. All computers have a realistic limit to the amount of input they can accept (a technical upper bound). Just because nothing in the world can store a yottabyte of data doesn't mean saying every algorithm is O(1) is useful! It's about applying the mathematics in a way that makes sense for the situation.
Here are two contexts for your example, one where it makes sense to call it O(1), and one where it does not.
"I decided I won't put strings of length more than 15 into my program, therefore it is O(1)". This is not a super useful interpretation of the runtime. The actual time is still strongly tied to the size of the string; a string of size 1 will run much faster than one of size 15 even if there is technically a constant bound. In other words, within the constraints of your problem there is still a strong correlation to n.
"My algorithm will process a list of n strings, each with maximum size 15". Here we have a different story; the runtime is dominated by having to run through the list! There's a point where n is so large that the time to process a single string doesn't change the correlation. Now it makes sense to consider the time to process a single string O(1), and therefore the time to process the whole list O(n)
That said, Big-O notation doesn't have to only use one variable! There are problems where upper bounds are intrinsic to the algorithm, but you wouldn't put a bound on the input arbitrarily. Instead, you can describe each dimension of your input as a different variable:
n = list length
s = maximum string length
=> O(n*s)
It depends.
If your algorithm's requirements would grow if larger inputs were provided, then the algorithmic complexity can (and should) be evaluated independently of the inputs. So iterating over all the elements of a list, array, string, etc., is O(n) in relation to the length of the input.
If your algorithm is tied to the limited input size, then that fact becomes part of your algorithmic complexity. For example, maybe your algorithm only iterates over the first 15 characters of the input string, regardless of how long it is. Or maybe your business case simply indicates that a larger input would be an indication of a bug in the calling code, so you opt to immediately exit with an error whenever the input size is larger than a fixed number. In those cases, the algorithm will have constant requirements as the input length tends toward very large numbers.
From Wikipedia
Big O notation is a mathematical notation that describes the limiting behavior of a function when the argument tends towards a particular value or infinity.
...
In computer science, big O notation is used to classify algorithms according to how their run time or space requirements grow as the input size grows.
In practice, almost all inputs have limits: you cannot input a number larger than what's representable by the numeric type, or a string that's larger than the available memory space. So it would be silly to say that any limits change an algorithm's asymptotic complexity. You could, in theory, use 15 as your asymptote (or "particular value"), and therefore use Big-O notation to define how an algorithm grows as the input approaches that size. There are some algorithms with such terrible complexity (or some execution environments with limited-enough resources) that this would be meaningful.
But if your argument (string length) does not tend toward a large enough value for some aspect of your algorithm's complexity to define the growth of its resource requirements, it's arguably not appropriate to use asymptotic notation at all.
NO!
The time complexity of an algorithm is independent of program constraints. Here is (a simple) way of thinking about it:
Say your algorithm iterates over the string and appends all consonants to a list.
Now, for iteration time complexity is O(n). This means that the time taken will increase roughly in proportion to the increase in the length of the string. (Time itself though would vary depending on the time taken by the if statement and Branch Prediction)
The fact that you know that the string is between 1 and 15 characters long will not change how the program runs, it merely tells you what to expect.
For example, knowing that your values are going to be less than 65000 you could store them in a 16-bit integer and not worry about Integer overflow.
Do problem constraints change the time complexity of algorithms?
No.
If I know for sure that the length of the string is less than, say, 15 characters ..."
We already know the length of the string is less than SIZE_MAX. Knowing an upper fixed bound for string length does not make the the time complexity O(1).
Time complexity remains O(n).
Big-O measures the complexity of algorithms, not of code. It means Big-O does not know the physical limitations of computers. A Big-O measure today will be the same in 1 million years when computers, and programmers alike, have evolved beyond recognition.
So restrictions imposed by today's computers are irrelevant for Big-O. Even though any loop is finite in code, that need not be the case in algorithmic terms. The loop may be finite or infinite. It is up to the programmer/Big-O analyst to decide. Only s/he knows which algorithm the code intends to implement. If the number of loop iterations is finite, the loop has a Big-O complexity of O(1) because there is no asymptotic growth with N. If, on the other hand, the number of loop iterations is infinite, the Big-O complexity is O(N) because there is an asymptotic growth with N.
The above is straight from the definition of Big-O complexity. There are no ifs or buts. The way the OP describes the loop makes it O(1).
A fundamental requirement of big-O notation is that parameters do not have an upper limit. Suppose performing an operation on N elements takes a time precisely equal to 3E24*N*N*N / (1E24+N*N*N) microseconds. For small values of N, the execution time would be proportional to N^3, but as N gets larger the N^3 term in the denominator would start to play an increasing role in the computation.
If N is 1, the time would be 3 microseconds.
If N is 1E3, the time would be about 3E33/1E24, i.e. 3.0E9.
If N is 1E6, the time would be about 3E42/1E24, i.e. 3.0E18
If N is 1E7, the time would be 3E45/1.001E24, i.e. ~2.997E21
If N is 1E8, the time would be about 3E48/2E24, i.e. 1.5E24
If N is 1E9, the time would be 3E51/1.001E27, i.e. ~2.997E24
If N is 1E10, the time would be about 3E54/1.000001E30, i.e. 2.999997E24
As N gets bigger, the time would continue to grow, but no matter how big N gets the time would always be less than 3.000E24 seconds. Thus, the time required for this algorithm would be O(1) because one could specify a constant k such that the time necessary to perform the computation with size N would be less than k.
For any practical value of N, the time required would be proportional to N^3, but from an O(N) standpoint the worst-case time requirement is constant. The fact that the time changes rapidly in response to small values of N is irrelevant to the "big picture" behaviour, which is what big-O notation measures.
It will be O(1) i.e. constant.
This is because for calculating time complexity or worst-case time complexity (to be precise), we think of the input as a huge chunk of data and the length of this data is assumed to be n.
Let us say, we do some maximum work C on each part of this input data, which we will consider as a constant.
In order to get the worst-case time complexity, we need to loop through each part of the input data i.e. we need to loop n times.
So, the time complexity will be:
n x C.
Since you fixed n to be less than 15 characters, n can also be assumed as a constant number.
Hence in this case:
n = constant and,
(maximum constant work done) = C = constant
So time complexity is n x C = constant x constant = constant i.e. O(1)
Edit
The reason why I have said n = constant and C = constant for this case, is because the time difference for doing calculations for smaller n will become so insignificant (compared to n being a very large number) for modern computers that we can assume it to be constant.
Otherwise, every function ever build will take some time, and we can't say things like:
lookup time is constant for hashmaps
I've seen coding problems that are similar to this:
int doSomething(String s)
Where it says in the problem description that s will contain at most one of every character, so s cannot be more than length 26. I think in this case, iterating over s would be constant time.
But I've also seen problems where inputs are constrained to a random large number, like 10^5, just to avoid stack overflows and other weird edge cases. If we are going to consider inputs that are constrained by constants to be constant complexity, shouldn't these inputs also be considered constant complexity?
But it doesn't make sense to me to consider s to be of O(n) complexity either, because there are many problems were people allocate char[26] arrays to hold every letter of the alphabet. How does if make sense to consider an input that we know will be less than or equal to 26 to be of greater complexity than an array of size 26?
The point of analyzing the complexity of algorithms is to estimate how long it will take to run it. If the problem you're trying to solve limits the maximum value of n to a constant, you can consider n to be a constant and you wouldn't be wrong. But would that be useful if you wanted to predict whether an algorithm that does 2^n operations will run in a few seconds for n = 26? On the other hand, if you had an algorithm that does n*m operations and m is at most 3, how useful would it be to include m in the complexity analysis?
Calculating complexity has focus on what is the most critical variable related to the running time. If the running time is dominant by the length of s, it is our main focus of analyzing complexity and that should be in bigO notation. And in that case, of course it's not a constant.
If the input is constrained to a large number like 10^5.
And if the algorithm is getting slower proportional to that input.
for example,
int sort(string s); //length of s is less than 10^5
In this case, depending on what sorting algorithm you use,
the running time will be proportional to the length of s
like O(n^2) or O(nlogn) if n is the length of s
In this case you cannot say it's constant because running time is very different as the length of s is changing.
But if the algorithm inside has nothing to do with the length of s, like it has constant calculation time, then you can say 10^5 constraint is just a constant.
Suppose I have an algorithm that runs in O(n) for every input of size n, but only after a pre-computation step of O(n^2) for that given size n. Is the algorithm considered O(n) still, with O(n^2) amortized? Or does big O only consider one "run" of the algorithm at size n, and so the pre-computation step is included in the notation, making the true notation O(n+n^2) or O(n^2)?
It's not uncommon to see this accounted for by explicitly separating out the costs into two different pieces. For example, in the range minimum query problem, it's common to see people talk about things like an ⟨O(n2), O(1)⟩-time solution to the problem, where the O(n2) denotes the precomputation cost and the O(1) denotes the lookup cost. You also see this with string algorithms sometimes: a suffix tree provides an O(m)-preprocessing-time, O(n+z)-query-time solution to string searching, while Aho-Corasick string matching offers an O(n)-preprocessing-time, O(m+z)-query-time solution.
The reason for doing so is that the tradeoffs involved here really depend on the use case. It lets you quantitatively measure how many queries you're going to have to make before the preprocessing time starts to be worth it.
People usually care about the total time to get things done when they are talking about complexity etc.
Thus, if getting to the result R requires you to perform steps A and B, then complexity(R) = complexity(A) + complexity(B). This works out to be O(n^2) in your particular example.
You have already noted that for O analysis, the fastest growing term dominates the overall complexity (or in other words, in a pipeline, the slowest module defines the throughput).
However, complexity analysis of A and B will be typically performed in isolation if they are disjoint.
In summary, it's the amount of time taken to get the results that counts, but you can (and usually do) reason about the individual steps independent of one another.
There are cases when you cannot only specify the slowest part of the pipeline. A simple example is BFS, with the complexity O(V + E). Since E = O(V^2), it may be tempting to write the complexity of BFS as O(E) (since E > V). However, that would be incorrect, since there can be a graph with no edges! In those cases, you will still need to iterate over all the vertices.
The point of O(...) notation is not to measure how fast the algorithm is working, because in many specific cases O(n) can be significantly slower than, say O(n^3). (Imagine the algorithm which runs in 10^100 n steps vs. the one which runs in n^3 / 2 steps.) If I tell you that my algorithm runs in O(n^2) time, it tells you nothing about how long it will take for n = 1000.
The point of O(...) is to specify how the algorithm behaves when the input size grows. If I tell you that my algorithm runs in O(n^2) time, and it takes 1 second to run for n = 500, then you'll expect rather 4 seconds to for n = 1000, not 1.5 and not 40.
So, to answer your question -- no, the algorithm will not be O(n), it will be O(n^2), because if I double the input size the time will be multiplied by 4, not by 2.
The time complexity of an algorithm is defined as the amount of time taken by it to run as a function of the length of the input.
If I have a simple for loop function in C, which runs for a given input n then:
the length of n is log n(number of bits required to represent it).
Since input is log n and the loop runs n times, the code runs exponentially many times in it's input length ( 2^(log n) = n))
C code:
int forfunction(unsigned int n){
unsigned int i=0;
for(;i<n;i++){
// do something ordinary
}
return 0;
}
This for loop being an example.
But we will never hear anyone say, that such a for loop program is exponential in it's input (the bits required to store n). Why is it so? The only difference I see is that this is a program and time complexity is defined for an algorithm. But even if it is, then why does this not have an impact/taken into account when we want to do a rough time complexity of a program?
EDIT:
Further clarification: I find it reasonable to claim it is exponential in it's input ( might be wrong =) ). If it so, then if a simple for loop is exponential, then what about other hard problems? Clearly this for loop is not a worry for anyone today. Why is it not? Why does this not have (much) real world implications when compared to other hard problems(EXP,NP-Hard,etc...)? Note: I am using hard the way it used to define NP-Hard problems.
Elaborating on #Anonymous's answer: The question you should be asking is "exponential in what?" Ultimately, whether this is exponential time depends on how n is presented to you.
If n is given to you as an explicit binary integer using O(log n) bits, then this function will run in pseudopolynomial time (technically exponential in the number of input bits, but polynomial in the numeric value of the input). This is why simple primality testing algorithms like trial division (divide n by all numbers from 2 up to √n and see if any of them are factors) technically run in exponential time even though they do run in time O(√n).
On the other hand, if n is given to you implicitly using O(n) bits (perhaps as the number of nodes in a graph given an adjacency matrix, or perhaps as the number of characters in a string given a string), then the runtime is polynomial because the input has at least linear size and linear work is done. This is why algorithms like DFS or BFS, which have runtimes of the form O(m + n), run in polynomial time: the number of bits in the input is Ω(m + n).
Hope this helps!
The amount of time a function takes is a function parameterised by something. Often it'll be the size of the input, but other times it's an explicit parameter, and it's up to you, when you're describing a function, to make it clear what you mean. Because it's often "obvious" what the parameterisation is from context, it's often omitted which leads to a lot of confusion when the parameterisation is not obvious to everyone.
When you add the word "complexity" then all that means is that instead of describing a function, you're saying it belongs to a particular class of functions. It doesn't obviate the need to say what the function is and what its argument is.
Techincally speaking the for loop or for that matter all linear programs are exponential in their inputs but this is not used to explain their runtime because for the simple reason that runtime is defined how it varies with change in input . Here in these problems it is considered the no of input bits is constant for example you might define the algorithm for only integer input so its input has always 32 bits so it is considered constant as even if value of n changes no of bits dont so constant terms cannot be used to define growth of algorithm hence omitted.
for example, say n = Integer.MAX_VALUE or 2^123 then O(log(n)) = 32 and 123 so a small integer. isn't it O(1) ?
what is the difference ? I think, the reason is O(1) is constant but O(log(n)) not. Any other ideas ?
If n is bounded above, then complexity classes involving n make no sense. There is no such thing as "in the limit as 2^123 approaches infinity", except in the old joke that "a pentagon approximates a circle, for sufficiently large values of 5".
Generally, when analysing the complexity of code, we pretend that the input size isn't bounded above by the resource limits of the machine, even though it is. This does lead to some slightly odd things going on around log n, since if n has to fit into a fixed-size int type, then log n has quite a small bound, so the bound is more likely to be useful/relevant.
So sometimes, we're analysing a slightly idealised version of the algorithm, because the actual code written cannot accept arbitrarily large input.
For example, your average quicksort formally uses Theta(log n) stack in the worst case, obviously so with the fairly common implementation that call-recurses on the "small" side of the partition and loop-recurses on the "big" side. But on a 32 bit machine you can arrange to in fact use a fixed-size array of about 240 bytes to store the "todo list", which might be less than some other function you've written based on an algorithm that formally has O(1) stack use. The morals are that implementation != algorithm, complexity doesn't tell you anything about small numbers, and any specific number is "small".
If you want to account for bounds, you could say that, for example, your code to sort an array is O(1) running time, because the array has to be below the size that fits in your PC's address space, and hence the time to sort it is bounded. However, you will fail your CS assignment if you do, and you won't be providing anyone with any useful information :-)
Obviously if you know that the input will always have a fixed number of elements, the algorithm will always run in constant time. Big-O notation is used to denote worse-case running time, which describes the limit when the number of elements grows infinitely large.
The difference is that n isn't fixed. The idea behind Big-O notation is to get an idea of how the size of the input effects the running time (or memory usage). So if an algorithm always takes the same amount of time, whether n = 1 or n = Integer.MAX_VALUE, we say it is O(1). If the algorithm takes a unit of time longer each time the input size doubles, then we say it is O(logn).
Edit: to answer your specific question on the difference between O(1) and O(logn), I'll give you an example. Let's say we want an algorithm that will find the min element in an unsorted array. One approach is to go through each element and keep track of the current min. Another approach is to sort the array and then return the first element.
The first algorithm is O(n), and the second algorithm is O(nlogn). So let's say we start with an array of 16 elements. The first algorithm will run in time 16, the second algorithm will run in time 16*4. If we increase it to 17, then it becomes 17 and 17*4. We might naively say that the second algorithm takes 4 times as long as the first algorithm (if we treat the logn component as constant).
But let's look at what happens when our array contains 2^32 elements. Now the first algorithm takes 2^32 time to complete, where our second algorithm takes 32*2^32 time to complete. It takes 32 times as long. Yes, it's a small difference, but it is still a difference. If the first algorithm takes 1 minute, the second algorithm will take over half an hour!
I think you will get a better idea if it is called O(n^0).
It is a scaling function depending on the input variable N. It is a function, not number, you should never assume any number for the variable N.
It is just like that you say that a function f(x) is 3 because f(100) = 3, it is wrong. It is a function, not any particular number. A constant function f(x) = 1 is still a function, it will never equal to another function g(x) = N, i.e. g(x)=f(x)
Its the growth rate that you want to look at. O(1) implies no growth at all. While O(logn) does have growth. Even though the growth is small it is still growth.
You’re not thinking big enough. Any algorithm that runs on a computer will either run forever or terminate after some small number of steps — since the computer is only a finite state machine, you cannot write algorithms that run for an arbitrary amount of time and then terminate. By that argument, Big-O notation is only theoretical and has no purpose in a real-life computer program. Even O(2^n) hits an upper limit at O(2^INT_MAX), which is equivalent to O(1).
Realistically, though, Big-O can help you out if you know the constant factors. Even if an algorithm has an upper bound of O(log n), and n can have 32 bits, that could mean the difference between a request taking 1 second and 32 seconds.
Big-O shows how running time (or memory, etc) changes as the size of problem changes.
When size of the problem gets 10 times bigger, an O(n) solution takes 10 times as long, an O(log(n)) solution takes a bit longer, and an O(1) solution takes the same time: O(1) means 'changes as fast as constant 1', but constants don't change.
Familiarize yourself with the big-O notation in a bit more detail.
There is a reason why you leave "O(n)" in, and consider to drop "O(log n)". They both are "constants": the former is less than 32, and the latter is less than 232. But you nevertheless have a natural feeling that you can't call O(n) O(1).
However, if log(n) < 32, it means that O(n*logn) algorithm works thirty two times slower than its O(n) version. Big enough to write "log*n"s?