I am asked to provide the asymptotic space and time complexity of the below algorithm in appropriate terms for an initial input number of arbitrary length (as distinct from a constant 12 digits).
1 for i = 2 to 12
2 if d[i] == d[i-1]
3 d[i] = (d[i] + 3) mod 10
This algorithm is apply for a number that has 12 digits, each digit is stored in the array d[] (so we have d[1], d[2],... d[12])
I figured out the time complexity is O(n) but how do I determine the space complexity?
Typically, to measure space complexity, you want to look at how much extra space is required to hold all the information necessary to execute the code. The challenge is that you can often reuse space across the lifetime of the code execution.
In this case, the code needs extra space to store
The values of temporary calculations in the loop,
The value of i, and
Miscellaneous data like the program counter, etc.
The last two of these take up O(1) space, since there's only one i and constant space for auxiliary data like the stack pointer, etc. So what about the first? Well, each iteration of the loop will need O(1) space for temporary variables, but notice that this space can get reused because after each loop iteration finishes, the space for those temporaries isn't needed anymore and can be reused in the next iteration. Therefore, the total space needed is just O(1).
(A note... are you sure the time complexity is O(n)? Notice that the number of iterations is a constant regardless of how large the array is.)
Hope this helps!
Related
There is a task where at first a sequence of numbers is given, and then several queries. In each request, an interval is indicated and it is necessary to say how many zeros are on it. There is a simple and efficient algorithm that uses prefix sums. But is there something better? Sorry for such a possibly stupid question from a newbie. (The improvement could be, for example, in speed, memory, or that the original sequence can be easily changed)
If you can afford linear space, you can create a second array of the same size where the i'th entry has the number of zeros up to (and including) i in the original array. Then the query to get zeroes from j to k inclusive is arr[k] - arr[j-1] if j > 0, or simply arr[k] if j is zero.
This may be the simple and efficient approach your allude to. Each query is O(1), and the up-front work is O(n) time & O(n) space.
Potential improvements:
If you don't care about preserving the input array, you could save space by overwriting it with the count of zeros array.
If the size of the input is large relative to the (avg query size) * (number of queries), your total time will be less if you don't precompute anything and instead just scan the portion of the input being queried on each query. Total time will be improved by query-time will be slower.
If the zeros are distributed according to some known formula, e.g., every even index has a zero, you can calculate the answer to a query in O(1) with no precomputation.
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!
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.
http://en.wikipedia.org/wiki/Dynamic_array#Performance
What exactly does it mean?
I thought inserting at the end would be O(n), as you'd have to allocate say, twice the space of the original array, and then move all the items to that location and finally insert the item. How is this O(1)?
Amortized O(1) efficiency means that the sum of the runtimes of n insertions will be O(n), even if any individual operation may take a lot longer.
You are absolutely correct that appending an element can take O(n) time because of the work required to copy everything over. However, because the array is doubled each time it is expanded, expensive doubling steps happen exponentially less and less frequently. As a result, the total work done in n inserts comes out to be O(n) rather than O(n2).
To elaborate: suppose you want to insert a total of n elements. The total amount of work done copying elements when resizing the vector will be at most
1 + 2 + 4 + 8 + ... + n ≤ 2n - 1
This is because first you copy one element, then twice that, then twice that, etc., and in the absolute worst case copy over all n elements. The sum of this geometric series works out to 2n - 1, so at most O(n) elements get moved across all copy steps. Since you do n inserts and only O(n) total work copying across all of them, the amortized efficiency is O(1) per operation. This doesn't say each operation takes O(1) time, but that n operations takes O(n) time total.
For a graphical intuition behind this, as well as a rationale for doubling the array versus just increasing it by a small amount, you might want to check out these lecture slides. The pictures toward the end might be very relevant.
Hope this helps!
Each reallocation in isolation is O(N), yes. But then on the next N insertions, you don't need to do anything. So the "average" cost per insertion is O(1). We say that "the cost is amortized across multiple operations".
This question already has answers here:
What is a plain English explanation of "Big O" notation?
(43 answers)
Closed 8 months ago.
I have seen this term "O(1) access time" used to mean "quickly" but I don't understand what it means. The other term that I see with it in the same context is "O(n) access time". Could someone please explain in a simple way what these terms mean?
See Also
What is Big O notation? Do you use it?
Big-O for Eight Year Olds?
You're going to want to read up on Order of complexity.
http://en.wikipedia.org/wiki/Big_O_notation
In short, O(1) means that it takes a constant time, like 14 nanoseconds, or three minutes no matter the amount of data in the set.
O(n) means it takes an amount of time linear with the size of the set, so a set twice the size will take twice the time. You probably don't want to put a million objects into one of these.
In essence, It means that it takes the same amount of time to look up a value in your collection whether you have a small number of items in your collection or very very many (within the constraints of your hardware)
O(n) would mean that the time it takes to look up an item is proportional to the number of items in the collection.
Typical examples of these are arrays, which can be accessed directly, regardless of their size, and linked lists, which must be traversed in order from the beginning to access a given item.
The other operation usually discussed is insert. A collection can be O(1) for access but O(n) for insert. In fact an array has exactly this behavior, because to insert an item in the middle, You would have to move each item to the right by copying it into the following slot.
O(1) means the time to access something is independent of the number of items in the collection.
O(N) would mean the time to access an item is a proportional to the number (N) of items in the collection.
Every answer currently responding to this question tells you that the O(1) means constant time (whatever it happens to measuring; could be runtime, number of operations, etc.). This is not accurate.
To say that runtime is O(1) means that there is a constant c such that the runtime is bounded above by c, independent of the input. For example, returning the first element of an array of n integers is O(1):
int firstElement(int *a, int n) {
return a[0];
}
But this function is O(1) too:
int identity(int i) {
if(i == 0) {
sleep(60 * 60 * 24 * 365);
}
return i;
}
The runtime here is bounded above by 1 year, but most of the time the runtime is on the order of nanoseconds.
To say that runtime is O(n) means that there is a constant c such that the runtime is bounded above by c * n, where n measures the size of the input. For example, finding the number of occurrences of a particular integer in an unsorted array of n integers by the following algorithm is O(n):
int count(int *a, int n, int item) {
int c = 0;
for(int i = 0; i < n; i++) {
if(a[i] == item) c++;
}
return c;
}
This is because we have to iterate through the array inspecting each element one at a time.
O(1) does not necessarily mean "quickly". It means that the time it takes is constant, and not based on the size of the input to the function. Constant could be fast or slow. O(n) means that the time the function takes will change in direct proportion to the size of the input to the function, denoted by n. Again, it could be fast or slow, but it will get slower as the size of n increases.
Here is a simple analogy;
Imagine you are downloading movies online, with O(1), if it takes 5 minutes to download one movie, it will still take the same time to download 20 movies. So it doesn't matter how many movies you are downloading, they will take the same time(5 minutes) whether it's one or 20 movies. A normal example of this analogy is when you go to a movie library, whether you are taking one movie or 5, you will simply just pick them at once. Hence spending the same time.
However, with O(n), if it takes 5 minutes to download one movie, it will take about 50 minutes to download 10 movies. So time is not constant or is somehow proportional to the number of movies you are downloading.
It's called the Big O notation, and describes the search time for various algorithms.
O(1) means that the worst-case run time is constant. For most situations it means that you don't actually need to search the collection, you can find what you are searching for right away.
"Big O notation" is a way to express the speed of algorithms. n is the amount of data the algorithm is working with. O(1) means that, no matter how much data, it will execute in constant time. O(n) means that it is proportional to the amount of data.
Basically, O(1) means its computation time is constant, while O(n) means it will depend lineally on the size of input - i.e. looping through an array has O(n) - just looping -, because it depends on the number of items, while calculating the maximum between two ordinary numbers has O(1).
Wikipedia might help as well: http://en.wikipedia.org/wiki/Computational_complexity_theory
O(1) always execute in the same time regardless of dataset n.
An example of O(1) would be an ArrayList accessing its element with index.
O(n) also known as Linear Order, the performance will grow linearly and in direct proportion to the size of the input data.
An example of O(n) would be an ArrayList insertion and deletion at random position. As each subsequent insertion/deletion at random position will cause the elements in the ArrayList to shift left right of its internal array in order to maintain its linear structure, not to mention about the creation of a new arrays and the copying of elements from the old to new array which takes up expensive processing time hence, detriment the performance.
The easiest way to differentiate O(1) and O(n) is comparing array and list.
For array, if you have the right index value, you can access the data instantly.
(If you don't know the index and have to loop through the array, then it won't be O(1) anymore)
For list, you always need to loop through it whether you know the index or not.
It means that access time is constant. Whether you're accessing from 100 or 100,000 records, the retrieval time will be the same.
In contrast, O(n) access time would indicate that the retrieval time is directly proportional to the number of records you're accessing from.
It means that the access takes constant time i.e. does not depend on the size of the dataset. O(n) means that the access will depend on the size of the dataset linearly.
The O is also known as big-O.
Introduction to Algorithms: Second Edition by Cormen, Leiserson, Rivest & Stein says on page 44 that
Since any constant is a degree-0
polynomial, we can express any
constant function as Theta(n^0), or
Theta(1). This latter notation is a
minor abuse, however, because it is
not clear what variable is tending to
infinity. We shall often use the
notation Theta(1) to mean either a
constant or a constant function with
respect to some variable. ... We
denote by O(g(n))... the set of
functions f(n) such that there exist
positive constants c and n0 such that
0 <= f(n) <= c*g(n) for all n >= n0.
... Note that f(n) = Theta(g(n))
implies f(n) = O(g(n)), since Theta
notation is stronger than O notation.
If an algorithm runs in O(1) time, it means that asymptotically doesn't depend upon any variable, meaning that there exists at least one positive constant that when multiplied by one is greater than the asymptotic complexity (~runtime) of the function for values of n above a certain amount. Technically, it's O(n^0) time.
O(1) means Random Access. In any Random Access Memory, the time taken to access any element at any location is the same. Here time can be any integer, but the only thing to remember is time taken to retrieve the element at (n-1)th or nth location will be same(ie constant).
Whereas O(n) is dependent on the size of n.
According to my perspective,
O(1) means time to execute one operation or instruction at a time is one, in time complexity analysis of algorithm for best case.