Here’s my favorite implementation of binary search. It’s useful not only for finding the element but also for finding the insertion index. Details about assumed sorting order (ascending or descending) and behavior with respect to equal elements are controlled by providing a corresponding predicate (e.g. { $0 < x } vs { $0 > x … Read more
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
This question was just asked again recently. Apart from Knuth’s quote that “Although the basic idea of binary search is comparatively straightforward, the details can be surprisingly tricky”, there is the staggering historical fact (see TAOCP, Volume 3, section 6.2.1) that binary search was first published in 1946 but the first published binary search without … Read more
For very small collections the difference is going to be negligible. At the low end of your range (500k items) you will start to see a difference if you’re doing lots of lookups. A binary search is going to be O(log n), whereas a hash lookup will be O(1), amortized. That’s not the same as … Read more
Start in the bottom-left corner of your matrix. Then go to the right until you find the exact number (done), or until you find a number that is bigger. Then you go upwards in the matrix until you find the exact number (done), or until you find a number that is too small. Then again … Read more
Since there are four proofs in the wikipedia article you referenced, it seems you aren’t looking for a mathematical explanation for the correspondence between the Catalan numbers and the permutations of a binary tree. So instead, here are two ways to try and intuitively visualise how the Catalan sequence (1, 2, 5, 14, 42, …) … Read more
I am a big fan of Java’s MappedByteBuffers for situations like this. It is blazing fast. Below is a snippet I put together for you that maps a buffer to the file, seeks to the middle, and then searches backwards to a newline character. This should be enough to get you going? I have similar … 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