It is certainly not possible with a plain singly-linked list. Sketch proof: to examine the last node of a singly-linked list, we must perform n-1 operations of following a “next” pointer [proof by induction on the fact that there is only one reference to the k+1th node, and it is in the kth node, and … Read more
A linear search looks down a list, one item at a time, without jumping. In complexity terms this is an O(n) search – the time taken to search the list gets bigger at the same rate as the list does. A binary search is when you start with the middle of a sorted list, and … Read more
One way of thinking about this problem is to think about doing a binary search over a transformed version of the array, where the array has been modified by applying the function f(x) = 1 if x > target 0 else Now, the goal is to find the very first place that this function takes … Read more
Well, as the list is sorted, all the entries you are interested in are contiguous. This means you need to find the first item equal to the found item, looking backwards from the index which was produced by the binary search. And the same about last item. You can simply go backwards from the found … Read more
It would be much better to work with a lower and upper indexes as Lasse V. Karlsen was suggesting in a comment to the question. This would be the code: def binary_search(array, target): lower = 0 upper = len(array) while lower < upper: # use < instead of <= x = lower + (upper – … Read more
I doubt there is a general purpose binary search method in .NET like that, except for the one being present in some base classes (but apparently not in the interfaces), so here’s my general purpose one. public static Int32 BinarySearchIndexOf<T>(this IList<T> list, T value, IComparer<T> comparer = null) { if (list == null) throw new … Read more
This post covers this famous bug in a lot of detail. As others have said it’s an overflow issue. The fix recommended on the link is as follows: int mid = low + ((high – low) / 2); // Alternatively int mid = (low + high) >>> 1; It is also probably worth mentioning that … Read more
Ensure that your array is sorted since this is the crux of a binary search. Any indexed/random-access data structure can be binary searched. So when you say using “just an array”, I would say arrays are the most basic/common data structure that a binary search is employed on. You can do it recursively (easiest) or … Read more
There is no such functions, but you can write a simple one using std::lower_bound, std::upper_bound or std::equal_range. A simple implementation could be template<class Iter, class T> Iter binary_find(Iter begin, Iter end, T val) { // Finds the lower bound in at most log(last – first) + 1 comparisons Iter i = std::lower_bound(begin, end, val); if … Read more
Here a more mathematical way of seeing it, though not really complicated. IMO much clearer as informal ones: The question is, how many times can you divide N by 2 until you have 1? This is essentially saying, do a binary search (half the elements) until you found it. In a formula this would be … Read more