I am having a hard time understanding what is O(1) space complexity. I understand that it means that the space required by the algorithm does not grow with the input or the size of the data on which we are using the algorithm. But what does it exactly mean?
If we use an algorithm on a linked list say 1->2->3->4, to traverse the list to reach "3" we declare a temporary pointer. And traverse the list until we reach 3. Does this mean we still have O(1) extra space? Or does it mean something completely different. I am sorry if this does not make sense at all. I am a bit confused.
To answer your question, if you have a traversal algorithm for traversing the list which allocate a single pointer to do so, the traversal algorithms is considered to be of O(1) space complexity. Additionally, let's say that traversal algorithm needs not 1 but 1000 pointers, the space complexity is still considered to be O(1).
However, if let's say for some reason the algorithm needs to allocate 'N' pointers when traversing a list of size N, i.e., it needs to allocate 3 pointers for traversing a list of 3 elements, 10 pointers for a list of 10 elements, 1000 pointers for a list of 1000 elements and so on, then the algorithm is considered to have a space complexity of O(N). This is true even when 'N' is very small, eg., N=1.
To summarise the two examples above, O(1) denotes constant space use: the algorithm allocates the same number of pointers irrespective to the list size. In contrast, O(N) denotes linear space use: the algorithm space use grows together with respect to the input size.
It is just the amount of memory used by a program. the amount of computer memory that is the main memory required by the algorithm to complete its execution with respect to the input size.
Space Complexity(s(P)) of an algorithm is the total space taken by the algorithm to complete its execution with respect to the input size. It includes both Constant space and Auxiliary space.
S(P) = Constant space + Auxiliary space
Constant space is the one that is fixed for that algorithm, generally equals to space used by input and local variables. Auxiliary Space is the extra/temporary space used by an algorithm.
Let's say I create some data structure with a fixed size, and no matter what I do to the data structure, it will always have the same fixed size. Operations performed on this data structure are therefore O(1).
An example, let's say I have an array of fixed size 100. Any operation I do, whether that is reading from the array or updating an element, that operation will be O(1) on the array. The array's size (and thus the amount of memory it's using) is not changing.
Another example, let's say I have a LinkedList to which I add elements to it. Every time I add an element to the LinkedList, that is a O(N) operation to the list because I am growing the amount of memory required to hold all of it's elements together.
Hope this helps!
Related
Consider the following pseudocode:
linked_list_node = ... //We have some linked list
while linked_list_node is not NULL //Iterate through it
node_copy = CopyNode(linked_list_node) //We allocate a new pointer and copy the node, which is O(1) space
... //Do something
DeleteAndFree(node_copy) //We free the memory allocated at the beginning of the loop
Next(linked_list_node) //Advance once
Let N be the size of our linked list
On the one hand
At each iteration of the loop, we used O(1) space, the loop is N iterations, which means that in total we allocated O(N) space
On the other hand
We never actually allocated N nodes at the same time, each time we allocated exactly one node, so, in theory, we only O(1) Space. In other words, if our machine only had 1 byte available in memory, it could allocate and delete that same byte over and over again, never running into a memory limit.
I found this question on stack overflow: What is O(1) space complexity?
From accepted answer:
However, if let's say for some reason the algorithm needs to allocate 'N' pointers when traversing a list of size N, ..., then the algorithm is considered to have a space complexity of O(N)
It seems like my algorithm doesn't satisfy this condition as I never actually use N different pointers at the same time, so it should be O(1) Space. However, it indeed requires N allocating operations, which could be why it is really O(N) Space
So, what is the space complexity of that and why?
This would be O(1) complexity. At each step you can say that your usage increases and decreases by 1, so net gain of 0 for each element. That said, this is kind of an odd solution, as you presumably could replace it with an implementation where you allocate a single node, and just copy each element into it. The two would be equivalent, and the latter is clearly O(1) space complexity.
I know there are other questions about the meaning of the "in-place" algorithm but my question is a bit different. I know it means that the algorithm changes the original input data instead of allocating new space for the output. But what I'm not sure about is whether the auxiliary memory counts. Namely:
if an algorithm allocates some additional memory in order to compute the result
if an algorithm has a non-constant number of recursive calls which take up additional space on the stack
In-place normally implies sub-linear additional space. This isn't necessarily part of the meaning of the term. It's just that an in-place algorithm that uses linear or greater space is not interesting. If you're going to allocate O(n) space to compute an output in the same space as the input, you could have equally easily produced the output in fresh memory and maintained the same memory bound. The value of computing in-place has been lost.
Wikipedia goes farther and says the amount of extra storage is constant. However, an algorithm (say mergesort) that uses log(n) additional space to write the output over the input is still in-place in usages I have seen.
I can't think of any in-place algorithm that doesn't need some additional memory. Whether an algorithm is "in-place" is characterized by the following:
in-place: To perform an algorithm on an input of size Θ(f(n)) using o(f(n)) extra space by mutating the input into the output.
Take for example an in-place implementation of the "Insertion Sort" sorting algorithm. The input is a list of numbers taking Θ(n) space. It takes Θ(n2) time to run in the worst case, but it only takes O(1) space. If you were to not do the sort in-place, you would be required to use at least Ω(n) space, because the output needs to be a list of n numbers.
I have seen that in most cases the time complexity is related to the space complexity and vice versa. For example in an array traversal:
for i=1 to length(v)
print (v[i])
endfor
Here it is easy to see that the algorithm complexity in terms of time is O(n), but it looks to me like the space complexity is also n (also represented as O(n)?).
My question: is it possible that an algorithm has different time complexity than space complexity?
The time and space complexities are not related to each other. They are used to describe how much space/time your algorithm takes based on the input.
For example when the algorithm has space complexity of:
O(1) - constant - the algorithm uses a fixed (small) amount of space which doesn't depend on the input. For every size of the input the algorithm will take the same (constant) amount of space. This is the case in your example as the input is not taken into account and what matters is the time/space of the print command.
O(n), O(n^2), O(log(n))... - these indicate that you create additional objects based on the length of your input. For example creating a copy of each object of v storing it in an array and printing it after that takes O(n) space as you create n additional objects.
In contrast the time complexity describes how much time your algorithm consumes based on the length of the input. Again:
O(1) - no matter how big is the input it always takes a constant time - for example only one instruction. Like
function(list l) {
print("i got a list");
}
O(n), O(n^2), O(log(n)) - again it's based on the length of the input. For example
function(list l) {
for (node in l) {
print(node);
}
}
Note that both last examples take O(1) space as you don't create anything. Compare them to
function(list l) {
list c;
for (node in l) {
c.add(node);
}
}
which takes O(n) space because you create a new list whose size depends on the size of the input in linear way.
Your example shows that time and space complexity might be different. It takes v.length * print.time to print all the elements. But the space is always the same - O(1) because you don't create additional objects. So, yes, it is possible that an algorithm has different time and space complexity, as they are not dependent on each other.
Time and Space complexity are different aspects of calculating the efficiency of an algorithm.
Time complexity deals with finding out how the computational time of
an algorithm changes with the change in size of the input.
On the other hand, space complexity deals with finding out how much
(extra)space would be required by the algorithm with change in the
input size.
To calculate time complexity of the algorithm the best way is to check if we increase in the size of the input, will the number of comparison(or computational steps) also increase and to calculate space complexity the best bet is to see additional memory requirement of the algorithm also changes with the change in the size of the input.
A good example could be of Bubble sort.
Lets say you tried to sort an array of 5 elements.
In the first pass you will compare 1st element with next 4 elements. In second pass you will compare 2nd element with next 3 elements and you will continue this procedure till you fully exhaust the list.
Now what will happen if you try to sort 10 elements. In this case you will start with comparing comparing 1st element with next 9 elements, then 2nd with next 8 elements and so on. In other words if you have N element array you will start of by comparing 1st element with N-1 elements, then 2nd element with N-2 elements and so on. This results in O(N^2) time complexity.
But what about size. When you sorted 5 element or 10 element array did you use any additional buffer or memory space. You might say Yes, I did use a temporary variable to make the swap. But did the number of variables changed when you increased the size of array from 5 to 10. No, Irrespective of what is the size of the input you will always use a single variable to do the swap. Well, this means that the size of the input has nothing to do with the additional space you will require resulting in O(1) or constant space complexity.
Now as an exercise for you, research about the time and space complexity of merge sort
First of all, the space complexity of this loop is O(1) (the input is customarily not included when calculating how much storage is required by an algorithm).
So the question that I have is if its possible that an algorithm has different time complexity from space complexity?
Yes, it is. In general, the time and the space complexity of an algorithm are not related to each other.
Sometimes one can be increased at the expense of the other. This is called space-time tradeoff.
There is a well know relation between time and space complexity.
First of all, time is an obvious bound to space consumption: in time t
you cannot reach more than O(t) memory cells. This is usually expressed
by the inclusion
DTime(f) ⊆ DSpace(f)
where DTime(f) and DSpace(f) are the set of languages
recognizable by a deterministic Turing machine in time
(respectively, space) O(f). That is to say that if a problem can
be solved in time O(f), then it can also be solved in space O(f).
Less evident is the fact that space provides a bound to time. Suppose
that, on an input of size n, you have at your disposal f(n) memory cells,
comprising registers, caches and everything. After having written these cells
in all possible ways you may eventually stop your computation,
since otherwise you would reenter a configuration you
already went through, starting to loop. Now, on a binary alphabet,
f(n) cells can be written in 2^f(n) different ways, that gives our
time upper bound: either the computation will stop within this bound,
or you may force termination, since the computation will never stop.
This is usually expressed in the inclusion
DSpace(f) ⊆ Dtime(2^(cf))
for some constant c. the reason of the constant c is that if L is in DSpace(f) you only
know that it will be recognized in Space O(f), while in the previous
reasoning, f was an actual bound.
The above relations are subsumed by stronger versions, involving
nondeterministic models of computation, that is the way they are
frequently stated in textbooks (see e.g. Theorem 7.4 in Computational
Complexity by Papadimitriou).
Yes, this is definitely possible. For example, sorting n real numbers requires O(n) space, but O(n log n) time. It is true that space complexity is always a lowerbound on time complexity, as the time to initialize the space is included in the running time.
Sometimes yes they are related, and sometimes no they are not related,
actually we sometimes use more space to get faster algorithms as in dynamic programming https://www.codechef.com/wiki/tutorial-dynamic-programming
dynamic programming uses memoization or bottom-up, the first technique use the memory to remember the repeated solutions so the algorithm needs not to recompute it rather just get them from a list of solutions. and the bottom-up approach start with the small solutions and build upon to reach the final solution.
Here two simple examples, one shows relation between time and space, and the other show no relation:
suppose we want to find the summation of all integers from 1 to a given n integer:
code1:
sum=0
for i=1 to n
sum=sum+1
print sum
This code used only 6 bytes from memory i=>2,n=>2 and sum=>2 bytes
therefore time complexity is O(n), while space complexity is O(1)
code2:
array a[n]
a[1]=1
for i=2 to n
a[i]=a[i-1]+i
print a[n]
This code used at least n*2 bytes from the memory for the array
therefore space complexity is O(n) and time complexity is also O(n)
The way in which the amount of storage space required by an algorithm varies with the size of the problem it is solving. Space complexity is normally expressed as an order of magnitude, e.g. O(N^2) means that if the size of the problem (N) doubles then four times as much working storage will be needed.
space complexity is the total amount of memory space used by an algorithm/program, including input value execution space. whereas the time complexity is the number of operations an algorithm performs to complete its task. These are two different concept, a single algorithm can of low time complexity but still can take up a lot of memory for example hashmaps take more memory than array but take less time.
I was wondering if there was a simple data structure that supports amortized log(n) lookup and insertion like a self balancing binary search tree but with constant memory overhead. (I don't really care about deleting elements).
One idea I had was to store everything in one contiguous block of memory divided into two contiguous blocks: an S part where all elements are sorted, and a U that isn't sorted.
To perform an insertion, we could add an element to U, and if the size of U exceeds log(size of S), then you sort the entire contiguous array (treat both S and U as one contiguous array), so that after the sort everything is in S and U is empty.
To perform lookup run binary search on S and just look through all of U.
However, I am having trouble calculating the amortized insertion time of my algorithm.
Ultimately I would just appreciate some reasonably simple algorithm/datastructure with desired properties, and some guarantee that it runs reasonably fast in amortized time.
Thank you!
If by constant amount of memory overhead you mean that for N elements stored in the data-structure the space consumption should be O(N), then any balanced tree will do -- in fact, any n-ary tree storing the elements in external leaves, where n > 1 and every external tree contains an element, has this property.
This follows from the fact that any tree graph with N nodes has N - 1 edges.
If by constant amount of memory overhead you mean that for N elements the space consumption should be N + O(1), then neither the balanced trees nor the hash tables have this property -- both will use k * N memory, where k > 1 due to extra node pointers in the case of trees and the load factor in the case of hash tables.
I find your approach interesting, but I do not think it will work even if you only sort U, and then merge the two sets in linear time. You would need to do a sort (O(logN * log(logN)) operations) after every logN updates, followed by an O(n) merging of S and U (note that so far nobody actually knows how to do this in linear time in place, that is, without an extra array).
The amortized insertion time would be O(n / logN). But you could maybe use your approach to achieve something close to O(√n) if you allow the size of U to grow to the square root of S.
Any hashtable will do that. The only tricky part about it is how you resolve conflicts - there are few ways of doing it, the other tricky part is correct hash computing.
See:
http://en.wikipedia.org/wiki/Hash_table
Inserting an element into a heap involves appending it to the end of the array and then propagating it upwards until it's in the "right spot" and satisfies the heap property, the operation of which is O(logn).
However, in C, for instance, calling realloc in order to resize the array for the new element can (and likely will) result in having to copy the entirety of the array to another location in memory, which is O(n) in the best and worst case, right?
Are heaps in C (or any language, for that matter) usually done with a fixed, pre-allocated size, or is the copy operation inconsequential enough to make a dynamically sized heap a viable choice (e.g, a binary heap to keep a quickly searchable list of items)?
A typical scheme is to double the size when you run out of room. This doubling--and the copying that goes with it--does indeed take O(n) time.
However, notice that you don't have to perform this doubling very often. If you average out the total cost of all the doubling over all the operations performed on the heap that did not involve doubling, then the cost is indeed inconsequential. (This kind of averaging is known as amortized analysis.)