It is reasonable to expect that you cannot do any better than O(N log N) in running time. However, the interesting part is to investigate whether you can sort it in-place, stably, its worst-case behavior and so on. Simon Tatham, of Putty fame, explains how to sort a linked list with merge sort. He concludes … Read more
This depends entirely on how smart you want the algorithm to be. For instance, here are some issues: cropped images vs. an uncropped image images with a text added vs. another without mirrored images The easiest and simplest algorithm I’ve seen for this is just to do the following steps to each image: scale to … Read more
There is good essay by Peter Norvig how to implement a spelling corrector. It’s basicly a brute force approach trying candidate strings with a given edit distance. (Here are some tips how you can improve the spelling corrector performance using a Bloom Filter and faster candidate hashing.) The requirements for a spell checker are weaker. … Read more
Starting from root node and moving downwards if you find any node that has either p or q as its direct child then it is the LCA. (edit – this should be if p or q is the node’s value, return it. Otherwise it will fail when one of p or q is a direct … Read more
Keep 2 stacks, let’s call them inbox and outbox. Enqueue: Push the new element onto inbox Dequeue: If outbox is empty, refill it by popping each element from inbox and pushing it onto outbox Pop and return the top element from outbox Using this method, each element will be in each stack exactly once – … Read more
As @IVlad noted in the comment to your question Yodaness problem asks you to count number of inversions and not minimal number of swaps. For example: L1 = [2,3,4,5] L2 = [2,5,4,3] The minimal number of swaps is one (swap 5 and 3 in L2 to get L1), but number of inversions is three: (5 … Read more
If there is no additional information, this question seems to be unsolveable, as this is the Element Distinctness Problem, which is unsolveable with the restrictions you provided, in the required time. you can allow: (1) more memory and use a hashtable / hashset and meet the O(n) time criteria. [iterate the array, check if an … Read more
Collision between circles is easy. Imagine there are two circles: C1 with center (x1,y1) and radius r1; C2 with center (x2,y2) and radius r2. Imagine there is a line running between those two center points. The distance from the center points to the edge of either circle is, by definition, equal to their respective radii. … Read more
You could put all elements of the first list into a hash set. Then, iterate the second one and, for each of its elements, check the hash to see if it exists in the first list. If so, output it as an element of the intersection.
An efficient way of computing this area is to use a sweep algorithm. Let us assume that we sweep a vertical line L(x) through the union of rectangles U: first of all, you need to build an event queue Q, which is, in this case, the ordered list of all x-coordinates (left and right) of … Read more