## Space Complexity of an array?

time complexity

space complexity python

time and space complexity

space complexity of for loop

space complexity of linear search

o(1) space complexity

space complexity of sorting algorithms

I have an array of size N, and N is <=200.

What would be the space complexity here.

**O(1)** or **(N)** - considering the constraint N.

Complexity is only relevant when you try to foresee the performances of your algorithm with various input. I don't think it has any meaning to just speak about the space-complexity of an array without any context.

If you'll always create an array of size N (hard-coded), it's **O(1)**, because no matter what input your algorithm crunches, the space taken by your array is the same.

If your N grows with the size of the input, it's **O(f(n))**, where f(n) is the relation between n (size of input) and N (size of array).

NOTE : the formulation O(...) is a mathematical symbol to represent magnitude without any regard to the multiplier (sorry for the lack of precision, I'm way over my math degree and never learned the english terms), so, if N is a constant, **O(N) = O(1)** (they have the exact same meaning).

And if I'm remembering it right, if *f* < C * *g* , O(f) = O(g). Thus, if N is < 200 , O(N) = O(200) = O(1)

**What does 'Space Complexity' mean?,** Space complexity of all these sorting algorithms is O(n) though. Maximum sum of even indexed elements obtained by right shift on an even sized subarray� Space Complexity of an algorithm is total space taken by the algorithm with respect to the input size. Space complexity includes both Auxiliary space and space used by input. For example, if we want to compare standard sorting algorithms on the basis of space, then Auxiliary Space would be a better criteria than Space Complexity. Merge Sort

Space complexity is usually only defined for Algorithms.

But let's be *crafty* and form an algorithm from your Question.

Input: N values, N <= 200 Algorithm: Store all values Output: None

Space complexity is the amount of memory you need to execute the algorithm, in relation to N.

When you store 1 number, you'll need one memory area. When you store 2 it doubles...
Your memory complexity is **O(n)** which means it grows linearly; Just like it would be for this algorithm:

Input: N values, N <= 18,446,744,073,709,551,616 (unsigned int 64). Algorithm: Store all values Output: None

**But 200 is a really small number, can't we just say O(1)?**

Let's get crafty again, because we can make this O(1):

Input: N values, N <= 200 Algorithm: Store all values in an array of size 200 Output: None

When you store 1 number you will need 200 memory areas. When you store 2 numbers you will need 200 memory areas. When you store 200 numbers you will need 200 memory areas. This means the memory is constant and independent from N. Thus the complexity is O(1).

It's important to note that O(1) doesn't mean the amount of memory you need is 1, it means that the amount of memory you need is not in any relation to N. And thus it doesn't grow when N grows.

**But what if my objects are 50GB Blu-ray Discs? O(1) should be very small but now it would be 10 Terabytes!**

At this point we may finally realize that we don't always need to use Big O notations. We could just say that we need to store 10 Terabyte of Data and buy some Hard Disks.
If your Teacher makes a fuss about whether you write O(1) for very small N or O(n), then he is a very bad teacher. The answer to this Question is neither going to change your life nor your career. *Big O Notation only makes sense for numbers that can grow incredible large.*

**Space Complexity of Algorithms,** In the above code, 4*n bytes of space is required for the array a[] elements. 4 bytes each for x , n , i and the return value. Hence the total memory requirement will� Space complexity is the amount of memory used by the algorithm (including the input values to the algorithm) to execute and produce the result. Sometime Auxiliary Space is confused with Space Complexity. But Auxiliary Space is the extra space or the temporary space used by the algorithm during it's execution.

It depends on the case of your problem.if you use only constant amount of memory (or space). So, space complexity is **O(1)**.

However, if you have some data structures like a 1D-array, designed to hold N elements, where N can vary from input-to-input, then the amount of memory required is dependent on N. When N is small, the space required is also small. When N is large, then space required is also large. So, there is a linear dependence on the space required and the input size. That is **O(N)** space.

Similarly, if you have a 2D-array of size NxN, then generally, the space required is **O(N^2)**.

Consider the following example of finding the minimum space required for the algorithm:

Scanner in = new Scanner(System.in); int n = in.nextInt(); int[] array = new int[n]; for(int i = 0; i < n; i++){ array[i] = in.nextInt(); } int minimum = array[0]; for(int i = 0; i < n; i++){ if (array[i] < minimum){ minimum = array[i]; } } System.out.println(minimum);

Here, we are having an array, whose size varies with `n`

. Total space requirement = `2 + N`

, where `2`

is for the variables `n`

and `minimum`

and `N`

is for the `array`

. So, the space complexity of this algorithm is **O(N)**.

I hope this was what you were looking for.

**How do you calculate spatial complexity? - general,** All I just found is this: Space complexity describes how much space your program needs. Since foo does not declare arrays, each level requires� Similarly, Space complexity of an algorithm quantifies the amount of space or memory taken by an algorithm to run as a function of the length of the input. Time and space complexity depends on lots of things like hardware, operating system, processors, etc. However, we don't consider any of these factors while analyzing the algorithm.

I think it will be O(1).space complexity is O(n) if the space size increase linearly with increase of n.But in your case the function is not dependent on n after 200; f(n)=a*n+b..

**Time and Space Complexity Tutorials & Notes,** We will only consider the execution time of an algorithm. Lets start with a simple example. Suppose you are given an array� Thus, time complexity of merge sort algorithm is T(n) = Θ(nlogn). Also Read-Master’s Theorem for Solving Recurrence Relations . Space Complexity Analysis- Merge sort uses additional memory for left and right sub arrays. Hence, total Θ(n) extra memory is needed. Properties- Some of the important properties of merge sort algorithm are-

I have an array of size N, and N is <= 200.

Ok, you have an array of size N. What about it? meaning, having some data stored does not have any meaning in terms of space complexity since there is no context like some code (algorithm) that uses this array (space). So you can't measure space complexity of nothing (no code to be run, just data sitting there).

Now, if you use this array in some context like a function that creates N times this input array where N <= length of this array, then you can measure how space grows in relation to run time (number of statements/lines executed) of this function.

What would be the space complexity here.

O(1) or (N) - considering the constraint N?

In this case, space complexity would be O(1) because there is no grows in run time since there is no code to be executed. just one piece of data (your array).

I hope this helps

**Big-O Algorithm Complexity Cheat Sheet (Know Thy Complexities ,** Data Structure, Time Complexity, Space Complexity. Average, Worst, Worst. Access, Search, Insertion, Deletion, Access, Search, Insertion, Deletion. Array, Θ( 1)� An array can be said as a Mountain Array if it shows the following properties: The length of the given array is should be greater than or equal to 3 LENGTH >=3 . There can be only one peak or largest element in the array.

**What is the generic formula to the space complexity of an N ,** Q: What is the generic formula to the space complexity of an N dimensional array Space complexity is the calculation of how much space is needed in memory to� Q: What is the generic formula to the space complexity of an N dimensional array Space complexity is the calculation of how much space is needed in memory to deal with a given algorithm as the size of working storage is managed (normally as a fact

**How do we calculate space complexity?,** To be more general, if we reserve any memory apart from the input, it counts towards the space complexity. For example declaring an array of size n would add� There is an array related problem, the requirement is that time complexity is O(n) and space complexity is O(1). If I use Arrays.sort(arr), and use a for loop to one pass loop, for example: public

**2.2.3. Time complexity, space complexity, and the O-notation,** Making predictions on the running time and space consumption of a program So approximately, the time complexity of the program “sort an array of n strings� Complexity Analysis for Next Greater Element in an Array Time Complexity . O(N) where n is the number of elements present in the array. Here we visit the value of the array exactly one time. This approach leads us to linear time complexity. Space Complexity. O(N) because the maximum size of the stack goes upto O(n). References

##### Comments

- its just N isn't it?
- Isn't the space-complexity relevant for algorithms ? What is your algorithm ? Given N, create an array of size N ? It's O(N). If it's inside another algorithm with an input of size n (independant of N) and you just need to create an Array of size N, it's O(1).
- @nafas yeah i thought the same. But N is not growing here, it's constant.
- @JeremyGrand yeah it makes sense.
- @nafas yeah. i agree. i think in this case, if the array size is fixed and its not growing in anywhere, the complexity should be O(1).
- I personally don't think this is true. the constant value K = 200, but N is variable, it can be small, or large (N < K constraint). Imagine K = 2^10000.... Would it still be O(1) because "there's a constant constraint?" No, right?, because N can be 0..2^10000, so the space it takes depends on the value of N, and it follows a linear pattern. Thus O(N)
- mate , I don't think its right, say in java Max Integer = 2^31 -1 , can you still consider N <= Max(Integer) to have O(1)? if its the case O(N) is completely irrelevant to anything
- Yeah, I was speaking on a mathematical standpoint. The thing is that in computing n (as the size of the input) is necessarily bounded. So, we actually need to look whether K is meaningful against n.
- You are right about the context though: If you make a function
`createArray(N){ return new Array(K) }`

, then yes, the space complexity of your code will be O(1). But`createArray(N){ return new Array(N) }`

is O(N) - @olivarra1 I guess OP's issue is that N grows with n (input size) but is bouded by 200 as a mean to forcibly prevent this part of the algorithm to take too much resources, which means when n is small enough, the array is O(n), but when n is big, it becomes O(1). I guess that's a O(1)