binary-search

I am not able to solve the following problem optimally nor finding an approach to do this anywhere. Given a N × M matrix in which each row is sorted, find the overall median of the matrix. Assume N*M is odd. For example, Matrix = [1, 3, 5] [2, 6, 9] [3, 6, 9] A = [1, 2, 3, 3, 5, 6, 6, 9, 9] Median is 5. So, we return 5. Note: No extra memory is allowed. Any help will be appreciated. sunkuet02 Consider the following process. If we consider the N*M matrix as 1-D array then the median is the element of 1+N*M/2 th element. Then consider x will be the median if x is an element of the matrix and

Most of us are familiar with the maximum sum subarray problem . I came across a variant of this problem which asks the programmer to output the maximum of all subarray sums modulo some number M. The naive approach to solve this variant would be to find all possible subarray sums (which would be of the order of N^2 where N is the size of the array). Of course, this is not good enough. The question is - how can we do better? Example: Let us consider the following array: 6 6 11 15 12 1 Let M = 13. In this case, subarray 6 6 (or 12 or 6 6 11 15 or 11 15 12) will yield maximum sum ( = 12 ). We can

What is the difference between binary search and binary search tree? Are they the same? Reading the internet it seems the second is only for trees (up to 2 children nodes) and binary search doesn't follow this rule. I didn't quite get it. 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 binary tree, or NULL) A binary search tree is a binary tree (i

function binarySearch(value) { var startIndex = 0, stopIndex = words.length - 1, middle = Math.floor((stopIndex + startIndex) / 2); while (words[middle] != value && startIndex < stopIndex) { // adjust search area if (value < words[middle]) { stopIndex = middle - 1; } else if (value > words[middle]) { startIndex = middle + 1; } // recalculate middle middle = Math.floor((stopIndex + startIndex) / 2); } } I am making a large list of words in the format of an array: e.g. ["a","ab","abc","b"] In alphabetical order. The problem I'm having is modifying my binary search algorithm to add the word in

This question already has an answer here: What is the behavior of integer division? 5 answers Is it guaranteed in C that 1/2 == 0 ? I need that for an implementation of binary search: /* * len is the array length * ary is an array of ptrs * f is a compare function * found is a ptr to the found element in the array * both i and offset are unsigned integers, used as indexes */ for(i = len/2; !found && i < len; i += offset) { res = f(c->ary[i]); if (res == 0) { found = c->ary[i]; } else { offset = (res < 0 ? -1 : 1) * (i/2); if (!offset) { i = len+1; } } } Yes, this is guaranteed. According to

What is the best way to search in a circular array? Example 1 array : 45 67 44 11 49 4 56 12 39 90 circular array 11, 49, 4, 56, 12, 39, 90, 45, 67 Is Binary search the right approach to start with? Binary search is only useful if the array is sorted. You didn't provide much info about the problem domain but one approach would be to use a set (or hash table). For every number you put in the array, also insert it in the set. Lookups in a set (or hash table) happen in constant time, so there's no "searching". When you remove an item from the array, also remove it from the set. If your circular

I'm having trouble combining these two algorithms together. I've been asked to modify Binary Search to return the index that an element should be inserted into an array. I've been then asked to implement a Binary Insertion Sort that uses my Binary Search to sort an array of randomly generated ints . My Binary Search works the way it's supposed to, returning the correct index whenever I test it alone. I wrote out Binary Insertion Sort to get a feel for how it works, and got that to work as well. As soon as I combine the two together, it breaks. I know I'm implementing them incorrectly together,

Given an element and an array, the Ruby#index method returns the position of the element in the array. I implemented my own index method using binary search expecting mine would outperform the built-in one. To my surprise, the built-in one ran approximately three times as fast as mine in an experiment. Any Rubyist knows the reason why? The built-in #index is not a binary search , it's just a simple iterative search. However, it is implemented in C rather than Ruby, so naturally it can be several orders of magnitude faster. 来源： https://stackoverflow.com/questions/7436155/rubyindex-method-vs

I have a dictionary with keys that are ints. I would like to get the largest key. I don't keep track of keys so they might be consecutive (e.g. 1,2,3,4,5,6) but might skip (1,3,4,5) although I doubt that makes any difference. Do I just use a binary search or is there a method? As far as I see you can hardly beat binary search for such a simple task - maybe you can halve it. If you have LINQ available, you should be able to do: myDictionary.Keys.Max(); A binary search would be the fastest but wouldn't work against a normal dictionary, since the Keys aren't stored in any particular order.