What is the worst case running time of declaring a 2D array? The 2d array is not strictly square. I have seen answers that state it is O(n) and I have also seen answers that state its O(n²).
In my mind, when declaring an object array such as this:
Object[][] gridArray = new Object[i][j];
//The number of elements n = i*j
The time complexity when declaring the array should be O(n) as it will just scale linearly depending on how many elements there will be. Am I missing something?
It depends on what you mean by n.
Some may define n as your i and see the j depended to that, for example a square n * n, for that definition you get O(n^2), O(i * i) or O(i * j) if j in O(i).
However you defined it as n = i * j. So yes, it is O(n) for your definition of n but that notation hides O(i * j) which may be more appropriate.
However note that it depends on whether your language actually initializes all those array entries at creation. Some languages don't do that!
I guess your code is Java, this language actually sets every entry to an initial value. In your case this would be null, thus it indeed creates i * j elements and sets them to null, yielding O(i * j).
Also take a look at Syntax for creating a two-dimensional array, which explains in detail how your code works.
Related
The question:
4-2 Parameter-passing costs
Throughout this book, we assume that parameter passing during procedure calls takes constant time, even if an N-element array is being passed. This assumption is valid in most systems because a pointer to the array is passed, not the array itself.
This problem examines the implications of three parameter-passing strategies:
An array is passed by pointer. Time Theta(1)
2. An array is passed by copying. Time Theta(N), where N is the size of the array.
An array is passed by copying only the subrange that might be accessed by the called procedure. Time Theta(q-p+1) if the subarray A[p..q] is passed.
a. Consider the recursive binary search algorithm for finding a number in a sorted array (see Exercise 2.3-5 ). Give recurrences for the worst-case running times of binary search when arrays are passed using each of the three methods above, and give good upper bounds on the solutions of the recurrences. Let N be the size of the original problem and n be the size of a subproblem.
b. Redo part (a) for the MERGE-SORT algorithm from Section 2.3.1.
I have trouble understanding how to solve the recurrence of case 2 where arrays are passed by copying for both algorithms.
Take the binary search algorithm of case 2 for example, the recurrence most answers give is T(n)=T(n / 2)+Theta(N). I have no trouble about that.
Here is an answer I find online that looks correct:
T(n)
= T(n/2) + cN
= 2cN + T(n/4)
= 3cN + T(n/8)
= Sigma(i = 0 to lgn - 1) (2^i cN / (2^i))
= cNlgn
= Theta(nlgn)
I have trouble understanding how the second last line can result in the last line's answer. Why not represent it in Theta(Nlgn)? And how can N become n in the Theta notation?
N and n feel somewhat connected. I have trouble understanding their connection and how that is applied in the solution.
Seems that N represents full array length and n is current chunk size.
But formulas really exploit initial value only, when you start from full length n=N - for example, look at T(n/4) for T(N/4), so n===N everywhere.
In this case you can replace n with N.
My answer will not be very theoretical, but maybe this "more empirical" approach will help you figure it out. Also check Master Theorem (analysis of algorithms) for easier analysis of recursive algorithms complexity
Let's solve the binary search first:
By pointer
O(logN) - acts like an iterative binary search, there will be logN calls each having O(1) complexity
Copying the whole array
O(NlogN) - since for each of the logN calls of the function we copy N elements
Copying only the accessed subarray
O(N) - this one is not that obvious, but can be easily seen that the copied subarrays will be of length, N, N/2, N/4, N/8 .... and summing all this terms will be 2*N
Now for the merge sort algorithm:
O(NlogN) - the same method as for a3, the iterated subarrays will be of lengths N, (N/2, N/2), (N/4, N/4, N/4, N/4) ...
O(N^2) - we make 2*N calls of the sorting function and each has a complexity of O(N) for copying the whole array
O(NlogN) - we will copy only the subarrays that we will iterate over, so the complexity will be as in b1
Let's suppose that I have an array M of n*m elements, so if I want to print its elements I can do something like:
for i=1 to m
for j=1 to n
print m[i,j]
next j
next i
I know that the print instruction is done in constant time, so in this case I would have an algorithm complexity of:
\sum_{i=1}^{m}\sum_{j=1}^{n}c=m.n.c
so I suppose is in the order of O(n)
But what happens if the array has the same number of rows and columns, I suppose the complexity is:
\sum_{i=1}^{n}\sum_{j=1}^{n}c=n.n.c
so it is of order O(n^{2})
are my assumptions correct?
I'm assuming that m and n are variables and not constants. In that case, the runtime of the algorithm should be O(mn), not O(n), since the runtime is directly proportional to the number of elements in the array. You derived this with a summation, but it might be easier to see by just looking at how much work is done per array element. Given this, you're correct that if m = n, the runtime is quadratic on n.
Hope this helps!
Say I have an array of integers A such that A[i] = j, and I want to "invert it"; that is, to create another array of integers B such that B[j] = i.
This is trivial to do procedurally in linear time in any language; here's a Python example:
def invert_procedurally(A):
B = [None] * (max(A) + 1)
for i, j in enumerate(A):
B[j] = i
return B
However, is there any way to do this functionally (as in functional programming, using map, reduce, or functions like those) in linear time?
The code might look something like this:
def invert_functionally(A):
# We can't modify variables in FP; we can only return a value
return map(???, A) # What goes here?
If this is not possible, what is the best (most efficient) alternative when doing functional programming?
In this context are arrays mutable or immutable? Generally I'd expect the mutable case to be about as straightforward as your Python implementation, perhaps aside from a few wrinkles with types. I'll assume you're more interested in the immutable scenario.
This operation inverts the indices and elements, so it's also important to know something about what constitutes valid array indices and impose those same constraints on the elements. Haskell has a class for index constraints called Ix. Any Ix type is ordered and has a range implementation to make an ordered list of indices ranging from one specified index to another. I think this Haskell implementation does what you want.
import Data.Array.IArray
invertArray :: (Ix x) => Array x x -> Array x x
invertArray arr = listArray (low,high) newElems
where oldElems = elems arr
newElems = indices arr
low = minimum oldElems
high = maximum oldElems
Under the hood listArray uses zipWith and range to associate indices in the specified range to the listed elements. That part ought to be linear time, and so is the one-time operation of extracting elements and indices from an array.
Whenever the sets of the input arrays indices and elements differ some elements will be undefined, which for better or worse blow up faster than Python's None. I believe you could overcome the undefined issue by implementing new Ix a instances over the Maybe monad, for instance.
Quick side-note: check out the invPerm example in the Haskell 98 Library Report. It does something similar to invertArray, but assumes up front that input array's elements are a permutation of its indices.
A solution needing mapand 3 operations:
toTuples views an the array as a list of tuples (i,e) where i is the index and e the element in the array at that index.
fromTuples creates and loads an array from a list of tuples.
swap which takes a tuple (a,b) and returns (b,a)
Hence the solution would be (in Haskellish notation):
invert = fromTuples . map swap . toTuples
(This question is inspired by deque::insert() at index?, I was surprised that it wasn't covered in my algorithm lecture and that I also didn't find it mentioned in another question here and even not in Wikipedia :). I think it might be of general interest and I will answer it myself ...)
Dynamic arrays are datastructures that allow addition of elements at the end in amortized constant time O(1) (by doubling the size of the allocated memory each time it needs to grow, see Amortized time of dynamic array for a short analysis).
However, insertion of a single element in the middle of the array takes linear time O(n), since in the worst case (i.e. insertion at first position) all other elements needs to be shifted by one.
If I want to insert k elements at a specific index in the array, the naive approach of performit the insert operation k times would thus lead to a complexity of O(n*k) and, if k=O(n), to a quadratic complexity of O(n²).
If I know k in advance, the solution is quite easy: Expand the array if neccessary (possibly reallocating space), shift the elements starting at the insertion point by k and simply copy the new elements.
But there might be situations, where I do not know the number of elements I want to insert in advance: For example I might get the elements from a stream-like interface, so I only get a flag when the last element is read.
Is there a way to insert multiple (k) elements, where k is not known in advance, into a dynamic array at consecutive positions in linear time?
In fact there is a way and it is quite simple:
First append all k elements at the end of the array. Since appending one element takes O(1) time, this will be done in O(k) time.
Second rotate the elements into place. If you want to insert the elements at position index. For this you need to rotate the subarray A[pos..n-1] by k positions to the right (or n-pos-k positions to the left, which is equivalent). Rotation can be done in linear time by use of a reverse operation as explained in Algorithm to rotate an array in linear time. Thus the time needed for rotation is O(n).
Therefore the total time for the algorithm is O(k)+O(n)=O(n+k). If the number of elements to be inserted is in the order of n (k=O(n)), you'll get O(n+n)=O(2n)=O(n) and thus linear time.
You could simply allocate a new array of length k+n and insert the desired elements linearly.
newArr = new T[k + n];
for (int i = 0; i < k + n; i++)
newArr[i] = i <= insertionIndex ? oldArr[i]
: i <= insertionIndex + k ? toInsert[i - insertionIndex - 1]
: oldArr[i - k];
return newArr;
Each iteration takes constant time, and it runs k+n times, thus O(k+n) (or, O(n) if you so like).
[Description] Given two integer arrays with the same length. Design an algorithm which can judge whether they're the same. The definition of "same" is that, if these two arrays were in sorted order, the elements in corresponding position should be the same.
[Example]
<1 2 3 4> = <3 1 2 4>
<1 2 3 4> != <3 4 1 1>
[Limitation] The algorithm should require constant extra space, and O(n) running time.
(Probably too complex for an interview question.)
(You can use O(N) time to check the min, max, sum, sumsq, etc. are equal first.)
Use no-extra-space radix sort to sort the two arrays in-place. O(N) time complexity, O(1) space.
Then compare them using the usual algorithm. O(N) time complexity, O(1) space.
(Provided (max − min) of the arrays is of O(Nk) with a finite k.)
You can try a probabilistic approach - convert the arrays into a number in some huge base B and mod by some prime P, for example sum B^a_i for all i mod some big-ish P. If they both come out to the same number, try again for as many primes as you want. If it's false at any attempts, then they are not correct. If they pass enough challenges, then they are equal, with high probability.
There's a trivial proof for B > N, P > biggest number. So there must be a challenge that cannot be met. This is actually the deterministic approach, though the complexity analysis might be more difficult, depending on how people view the complexity in terms of the size of the input (as opposed to just the number of elements).
I claim that: Unless the range of input is specified, then it is IMPOSSIBLE to solve in onstant extra space, and O(n) running time.
I will be happy to be proven wrong, so that I can learn something new.
Insert all elements from the first array into a hashtable
Try to insert all elements from the second array into the same hashtable - for each insert to element should already be there
Ok, this is not with constant extra space, but the best I could come up at the moment:-). Are there any other constraints imposed on the question, like for example to biggest integer that may be included in the array?
A few answers are basically correct, even though they don't look like it. The hash table approach (for one example) has an upper limit based on the range of the type involved rather than the number of elements in the arrays. At least by by most definitions, that makes the (upper limit on) the space a constant, although the constant may be quite large.
In theory, you could change that from an upper limit to a true constant amount of space. Just for example, if you were working in C or C++, and it was an array of char, you could use something like:
size_t counts[UCHAR_MAX];
Since UCHAR_MAX is a constant, the amount of space used by the array is also a constant.
Edit: I'd note for the record that a bound on the ranges/sizes of items involved is implicit in nearly all descriptions of algorithmic complexity. Just for example, we all "know" that Quicksort is an O(N log N) algorithm. That's only true, however, if we assume that comparing and swapping the items being sorted takes constant time, which can only be true if we bound the range. If the range of items involved is large enough that we can no longer treat a comparison or a swap as taking constant time, then its complexity would become something like O(N log N log R), were R is the range, so log R approximates the number of bits necessary to represent an item.
Is this a trick question? If the authors assumed integers to be within a given range (2^32 etc.) then "extra constant space" might simply be an array of size 2^32 in which you count the occurrences in both lists.
If the integers are unranged, it cannot be done.
You could add each element into a hashmap<Integer, Integer>, with the following rules: Array A is the adder, array B is the remover. When inserting from Array A, if the key does not exist, insert it with a value of 1. If the key exists, increment the value (keep a count). When removing, if the key exists and is greater than 1, reduce it by 1. If the key exists and is 1, remove the element.
Run through array A followed by array B using the rules above. If at any time during the removal phase array B does not find an element, you can immediately return false. If after both the adder and remover are finished the hashmap is empty, the arrays are equivalent.
Edit: The size of the hashtable will be equal to the number of distinct values in the array does this fit the definition of constant space?
I imagine the solution will require some sort of transformation that is both associative and commutative and guarantees a unique result for a unique set of inputs. However I'm not sure if that even exists.
public static boolean match(int[] array1, int[] array2) {
int x, y = 0;
for(x = 0; x < array1.length; x++) {
y = x;
while(array1[x] != array2[y]) {
if (y + 1 == array1.length)
return false;
y++;
}
int swap = array2[x];
array2[x] = array2[y];
array2[y] = swap;
}
return true;
}
For each array, Use Counting sort technique to build the count of number of elements less than or equal to a particular element . Then compare the two built auxillary arrays at every index, if they r equal arrays r equal else they r not . COunting sort requires O(n) and array comparison at every index is again O(n) so totally its O(n) and the space required is equal to the size of two arrays . Here is a link to counting sort http://en.wikipedia.org/wiki/Counting_sort.
given int are in the range -n..+n a simple way to check for equity may be the following (pseudo code):
// a & b are the array
accumulator = 0
arraysize = size(a)
for(i=0 ; i < arraysize; ++i) {
accumulator = accumulator + a[i] - b[i]
if abs(accumulator) > ((arraysize - i) * n) { return FALSE }
}
return (accumulator == 0)
accumulator must be able to store integer with range = +- arraysize * n
How 'bout this - XOR all the numbers in both the arrays. If the result is 0, you got a match.