binary-search

I am looking for a built-in Ruby method that has the same functionality as index but uses a binary search algorithm, and thus requires a pre-sorted array. I know I could write my own implementation, but according to " Ruby#index Method VS Binary Search ", the built-in simple iterative search used by index is faster than a pure-Ruby version of binary search, since the built-in method is written in C. Does Ruby provide any built-in methods that do binary search? Marc-André Lafortune Ruby 2.0 introduced Array#bsearch and Range#bsearch . For Ruby 1.9, you should look into the bsearch and binary

Recently I came across one interesting question on linked list. Sorted singly linked list is given and we have to search one element from this list. Time complexity should not be more than O(log n) . This seems that we need to apply binary search on this linked list. How? As linked list does not provide random access if we try to apply binary search algorithm it will reach O(n) as we need to find length of the list and go to the middle. Any ideas? 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

I've heard that it's possible to implement binary search over a doubly-linked list in O(n) time. Accessing a random element of a doubly-linked list takes O(n) time, and binary search accesses O(log n) different elements, so shouldn't the runtime be O(n log n) instead? templatetypedef It's technically correct to say that the runtime of binary search on a doubly-linked list is O(n log n), but that's not a tight upper bound. Using a slightly better implementation of binary search and a more clever analysis, it's possible to get binary search to run in time O(n). The basic idea behind binary

Given an array of integers, find the local minima. An element A[i] is defined as a local minimum if A[i-1] > A[i] and A[i] < A[i+1] where i = 1...n-2. In case of boundary elements, the number has to be just smaller than its adjacent number. I know if there is only one local minimum, then we can solve with modified binary search. But if it is known that there exist multiple local minima in the array, can it be solved in O(log n) time? templatetypedef If the array elements are not guaranteed to be distinct, then it's not possible to do this in O(log n) time. The reason for this is the following:

What is the difference between Linear search and Binary search? 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 see whether that's greater than or less than the value you're looking for, which determines whether the value is in the first or second half of the list. Jump to the half way through the sublist, and compare again etc. This is pretty much how humans typically look

In a general binary search, we are looking for a value which appears in the array. Sometimes, however, we need to find the first element which is either greater or less than a target. Here is my ugly, incomplete solution: // Assume all elements are positive, i.e., greater than zero int bs (int[] a, int t) { int s = 0, e = a.length; int firstlarge = 1 << 30; int firstlargeindex = -1; while (s < e) { int m = (s + e) / 2; if (a[m] > t) { // how can I know a[m] is the first larger than if(a[m] < firstlarge) { firstlarge = a[m]; firstlargeindex = m; } e = m - 1; } else if (a[m] < /* something */) {

String[] sortedArray = new String[]{"Quality", "Name", "Testing", "Package"}; // Search for the word "cat" int index = Arrays.binarySearch(sortedArray, "Quality"); I always get -3 . Problem is in "Name" . Why I can not have "Name" in my array? Any idea? In order to use binarySearch , you will need to sort the array yourself first: String[] sortedArray = new String[]{"Quality", "Name", "Testing", "Package"}; java.util.Arrays.sort(sortedArray); int index = Arrays.binarySearch(sortedArray, "Quality"); Jeffrey The array is must be sorted. From Javadoc of binarySearch(): The range must be sorted