It will have to go through all the elements in the array to do this. Array is a unique data structure where you have to specify a size when you initialize it. Order would be the source array’s size or in Big O terms its O(length). Infact this happens internally in an ArrayList. ArrayList wraps … Read more
I would say it is θ[r (n choose r)], the n choose r part is the number of times the generator has to yield and also the number of times the outer while iterates. In each iteration at least the output tuple of length r needs to be generated, which gives the additional factor r. … Read more
MergeSort time Complexity is O(nlgn) which is a fundamental knowledge. Merge Sort space complexity will always be O(n) including with arrays. If you draw the space tree out, it will seem as though the space complexity is O(nlgn). However, as the code is a Depth First code, you will always only be expanding along one … Read more
You could use a heap queue; it can give you the K largest or smallest numbers out of a list of size N in O(NlogK) time. The Python standard library includes the heapq module, complete with a heapq.nsmallest() function ready implemented: import heapq k_smallest = heapq.nsmallest(k, input_list) Internally, this creates a heap of size K … Read more
Naive base-conversion as you described takes quadratic time; you do about n bigint-by-smallint divisions, most of which take time linear in the size of the n-bit bigint. You can do base conversion in O(M(n) log(n)) time, however, by picking a power of target-base that’s roughly the square root of the to-be-converted number, doing divide-and-remainder by … Read more
The answer depends strongly on your implementation. For the example you gave there are several possible solutions and I would say that the naive way to implement a solution has better complexity when implemented iterative. Here are the two implementations: int iterative_fib(int n) { if (n <= 2) { return 1; } int a = … Read more
This is a very interesting question! I think it depends on the following aspects: accessing single row by index (index is sorted and unique) should have runtime O(m) where m << n_rows accessing single row by index (index is NOT unique and is NOT sorted) should have runtime O(n_rows) accessing single row by index (index … Read more
The following is even more efficient: int findn(int num) { if ( num < 10 ) return 1; if ( num < 100 ) return 2; //continue until max int } You could optimize this even further by doing a binary search, but that would be overkill.
The time complexity is O(N) on average, O(NM) worst case (N being the length of the longer string, M, the shorter string you search for). As of Python 3.10, heuristics are used to lower the worst-case scenario to O(N + M) by switching algorithms. The same algorithm is used for str.index(), str.find(), str.__contains__() (the in … Read more
Yes, you are right, it is O(n) where n – length of list. Look here for more information: https://www.ics.uci.edu/~pattis/ICS-33/lectures/complexitypython.txt