In a sorted container, the last element that is less than or equivalent to x, is the element before the first element that is greater than x. Thus you can call std::upper_bound, and decrement the returned iterator once. (Before decrementing, you must of course check that it is not the begin iterator; if it is, … Read more
Although this question is 5 years old, but there is a great article in googleblog which explains the problem and the solution in detail which is worth to share. It’s needed to mention that in current implementation of binary search in Java mid = lo + (hi – lo) / 2 calculation is not used, … Read more
You can do better than O(log n) if the values are integers, in which case the best worst-case running time you can achieve, in terms of n, is O(sqrt(log n)). Otherwise, there is no way to beat O(log n) unless there are patterns in the input sequence. There are two approaches used to beat O(log … Read more
Ruby 2.0 introduced Array#bsearch and Range#bsearch. For Ruby 1.9, you should look into the bsearch and binary_search gems. Other possibility is to use a different collection than an array, like using rbtree bsearch is in my backports gem, but this is a pure Ruby version, so quite a bit slower. Note that a pure Ruby … Read more
Binary Search Trees A node in a binary tree is a data structure that has an element, and a reference to two other binary trees, typically called the left and right subtrees. I.e., a node presents an interface like this: Node: element (an element of some type) left (a binary tree, or NULL) right (a … Read more
Binary search is used everywhere. Take any sorted collection from any language library (Java, .NET, C++ STL and so on) and they all will use (or have the option to use) binary search to find values. While true that you have to implement it rarely, you still have to understand the principles behind it to … Read more
We can do this as follow: Maintaining an array sum which at index ith, it contains the modulus sum from 0 to ith. For each index ith, we need to find the maximum sub sum that end at this index: For each subarray (start + 1 , i ), we know that the mod sum … 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
What you need is a way to look up some data given a key. With the key being an unsigned int, this gives you several possibilities. Of course, you could use a std::map: typedef std::map<unsigned int, record_t> my_records; However, there’s other possibilities as well. For example, it’s quite likely that a hash map would be … Read more
You want either lower_bound, upper_bound, or equal_range.