binary-search

This earlier question talks about doing binary search over a doubly-linked list in O(n) time. The algorithm in that answer work as follows: Go to the middle of the list to do the first comparison. If it's equal to the element we're looking for, we're done. If it's bigger than the element we're looking for, walk backwards halfway to the start and repeat. If it's smaller than the element we're looking for, walk forwards halfway to the start and repeat. This works perfectly well for a doubly -linked list because it's possible to move both forwards and backwards, but this algorithm wouldn't work

I use standard binary search to quickly return a single object in a sorted list (with respect to a sortable property). Now I need to modify the search so that ALL matching list entries are returned. How should I best do this? 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 index, but this way the solution may be as slow as O(n) if there are a lot

What is the point of the one-comparison-per-iteration binary search? And can you explain how it works? There are two reasons to binary search with one comparison per iteration. The less important is performance. Detecting an exact match early using two comparisons per iteration saves an average one iteration of the loop, whereas (assuming comparisons involve significant work) binary searching with one comparison per iteration almost halves the work done per iteration. Binary searching an array of integers, it probably makes little difference either way. Even with a fairly expensive comparison,

I have a sorted array and want to do binary search on it. So I'm asking if something is already available in Swift library like sort etc.? Or is there a type independend version available? Of course I could write it by my own, but I like to avoid reinventing the wheel again. Here's a generic way to use binary search: func binarySearch<T:Comparable>(inputArr:Array<T>, searchItem: T) -> Int? { var lowerIndex = 0; var upperIndex = inputArr.count - 1 while (true) { let currentIndex = (lowerIndex + upperIndex)/2 if(inputArr[currentIndex] == searchItem) { return currentIndex } else if (lowerIndex >

This question already has an answer here: Finding multiple entries with binary search 12 answers I've been tasked with creating a method that will print all the indices where value x is found in a sorted array. I understand that if we just scanned through the array from 0 to N (length of array) it would have a running time of O(n) worst case. Since the array that will be passed into the method will be sorted, I'm assuming that I can take advantage of using a Binary Search since this will be O(log n). However, this only works if the array has unique values. Since the Binary Search will finish

We want to search for a given element in a circular sorted array in complexity not greater than O(log n) . Example: Search for 13 in {5,9,13,1,3} . My idea was to convert the circular array into a regular sorted array then do a binary search on the resulting array, but my problem was the algorithm I came up was stupid that it takes O(n) in the worst case: for(i = 1; i < a.length; i++){ if (a[i] < a[i-1]){ minIndex = i; break; } } then the corresponding index of ith element will be determined from the following relation: (i + minInex - 1) % a.length it is clear that my conversion (from circular

I was reading an algorithms book which had the following algorithm for binary search: public class BinSearch { static int search ( int [ ] A, int K ) { int l = 0 ; int u = A. length −1; int m; while (l <= u ) { m = (l+u) /2; if (A[m] < K) { l = m + 1 ; } else if (A[m] == K) { return m; } else { u = m−1; } } return −1; } } The author says "The error is in the assignment m = (l+u)/2; it can lead to overflow and should be replaced by m = l + (u-l)/2 ." I can't see how that would cause an overflow. When I run the algorithm in my mind for a few different inputs, I don't see the mid's value going

Does anyone know a faster-than-linear algorithm for finding a duplicate in a sequential list of numbers? I'm working in Java now but any language or psuedo-code is fine. For example, given this int[] input: 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 7 | 8 | 9 Output would be either index or value '7'. I know the obvious traversal at O(n) linear time, but I'm trying to see if this is possible via binary search at O(log n) time. If you assume the numbers must start at 0 and be increasing by 1 you can compare the middle to the index. If the middle is the same go higher, if the middle is not, go lower. This