The binary search algorithm depends on equal (or direct) access time for any selected element of a list. A linked list, however, does not provide that, so a binary search is inappropriate for a linked list.
the major limitation of binary search is that there is a need of sorted array to perform binary search operation. if array is not sorted the output is either not correct or may be after a long number of steps and according to data structure the output should come in minimum number of steps.
All lists are linked lists; there is no such thing as a separate "sorted list". There are algorithms that can sort a list, of course, but they all work on linked lists.
It isn't. In car parking the only thing we need to keep track of is tickets. When we issue a new ticket we place it at the end of the array. When we extract a (paid) ticket, we simply move the last ticket into its place. This allows constant-time insertion and extraction. The only cost is that we must perform a linear search in order to locate a ticket for extraction, but we can perform linear searches across an array much quicker than we can with a linked-list because we don't have the additional cost of dereferencing individual nodes. Note that even if the list were in sorted order, we cannot use a binary search because lists do not allow constant-time random access. Arrays do allow efficient binary searches but we must maintain sorted order to do so and that may well cost more than a simple linear search. Given that linear search must be performed regardless of which container we use, an array is the better choice of the two.
You compare items in a linked list by searching for them. Iterate through the list, comparing elements with the search key. If you encounter end-of-list, then the key is not found, otherwise you have found the element desired. Note that this is a half linear search. Statistically, if an element is to be found, it will be found at the halfway point, assuming uniformly random distribution of data. In the worst case, if the element is not found, it will always take a full search to prove that. You could keep the linked list in order, by inserting each element before the element that has higher key value. This would reduce search time to half, because searching would stop with the element with higher key value. Searching is not a very efficient use of a linked-list. It would be better to use some kind of ordered list, perhaps a dynamic array with binary search, or a balanced binary tree, which has similar search performance but one with the most cost to design and implement. Linked-lists are better for keeping elements in the order they were encountered or inserted, such as processing tokens in a compiler. Sorry, but every solution has its tradeoffs.
Searching is slower for linked-lists rather than (ordered) trees because you need to iterate through each element serially, whereas for an ordered (binary) tree you use a "divide and conquer", otherwise known as binary search, method. Think of it this way. Think of a number between 1 and 128. Perhaps the number is 97. With an ordered list, it would take 97 comparisons to find the item. With an unordered list, it would take an average of 64 comparisons - if the order was random. With a binary tree, i.e. a binary search, you only need 7 comparisons. In fact, you never need more than 7 comparisions for a tree size of 128, and on average, you would use less than that.
There is no such thing. There are binary trees and linked lists.
Structures are data structures, which includes arrays, linked-lists, doubly-linked lists, circular lists, trees, binary trees and balanced binary trees, amongst many others. They are simply frameworks that are used to represent and manipulate data. For instance, a self-balancing binary tree is often used to automatically sort data as it is input, allowing for fast search and retrieval of that data. Lists are typically used for unsorted queues and stacks while arrays are typically used for high-speed random access.
the major limitation of binary search is that there is a need of sorted array to perform binary search operation. if array is not sorted the output is either not correct or may be after a long number of steps and according to data structure the output should come in minimum number of steps.
Binary trees main advantages are in searching and sorting. Algorithms for searching binary trees (Binary search trees) worst case ie. least efficient is when all the data is in a chain with no sub trees (like a linked list).So for data stored in a binary tree, at worst it will be as effective for searching and sorting as a linked list, stack etc. At best it will be much faster.The only disadvantage I can think of is reordering the tree on removal of an element is a little more complex than in say a list or stack. But still quite easy to implement.
No. Binary search tree will take less time to delete or insert an element. While deleting from list, more time will be required to search for the element to be deleted but BST will work faster if the no. of elements are more. Plus it depends upon which element is to be deleted and where the element is to be added. But the average time to perform these operations will be less in BST as compared to lists
1. Which field search button is used for long lists and provides a dialog box from which you select one or more criteria to perform a search for possible field entries?
All lists are linked lists; there is no such thing as a separate "sorted list". There are algorithms that can sort a list, of course, but they all work on linked lists.
The primary advantage of a binary search algorithm is that searching a sequence can be achieved in logarithmic time. The primary disadvantages are that the data must be in sorted order. Arrays are the ideal container for binary searches as they provide constant-time random-access and are therefore trivial to both sort and search efficiently. Sorted binary trees can also be used, however for optimal performance they must be totally balanced (e.g., red/black binary tree). Constant-time random-access is not a requirement of binary trees, however the cost of maintaining balance during construction of the tree has to be taken into account. With a linear search, we start at one end of the sequence and traverse through the sequence one element at a time until we find the value we're looking for, or we reach the element one-past-the-end of the sequence (in which case the element we're looking for does not exist). For a sequence of n elements, the worst case is O(n). Linear search is ideal for forward lists (singly-linked lists) and lists (doubly-linked lists) as neither provides nor requires constant-time random-access. With binary search, we locate the middle element in the sequence. If that's not the value we are looking for, we can easily determine which half of the sequence contains our value because the elements are in sorted order. So we eliminate the other half and repeat the algorithm with the remaining half. As such, each failure to find the value reduces the number of elements to be searched by half (plus the middle element). If there are no elements in the remaining half then the value does not exist. The worst case is therefore O(log n).
Calbar is an excellent source to perform a California attorney search. This database lists attorneys currently practicing in California. You can search under California attorney or California State Bar for a list of practicing attorneys in that state.
The primary advantage of a binary search algorithm is that searching a sequence can be achieved in logarithmic time. The primary disadvantages are that the data must be in sorted order. Arrays are the ideal container for binary searches as they provide constant-time random-access and are therefore trivial to both sort and search efficiently. Sorted binary trees can also be used, however for optimal performance they must be totally balanced (e.g., red/black binary tree). Constant-time random-access is not a requirement of binary trees, however the cost of maintaining balance during construction of the tree has to be taken into account. With a linear search, we start at one end of the sequence and traverse through the sequence one element at a time until we find the value we're looking for, or we reach the element one-past-the-end of the sequence (in which case the element we're looking for does not exist). For a sequence of n elements, the worst case is O(n). Linear search is ideal for forward lists (singly-linked lists) and lists (doubly-linked lists) as neither provides nor requires constant-time random-access. With binary search, we locate the middle element in the sequence. If that's not the value we are looking for, we can easily determine which half of the sequence contains our value because the elements are in sorted order. So we eliminate the other half and repeat the algorithm with the remaining half. As such, each failure to find the value reduces the number of elements to be searched by half (plus the middle element). If there are no elements in the remaining half then the value does not exist. The worst case is therefore O(log n).
It isn't. In car parking the only thing we need to keep track of is tickets. When we issue a new ticket we place it at the end of the array. When we extract a (paid) ticket, we simply move the last ticket into its place. This allows constant-time insertion and extraction. The only cost is that we must perform a linear search in order to locate a ticket for extraction, but we can perform linear searches across an array much quicker than we can with a linked-list because we don't have the additional cost of dereferencing individual nodes. Note that even if the list were in sorted order, we cannot use a binary search because lists do not allow constant-time random access. Arrays do allow efficient binary searches but we must maintain sorted order to do so and that may well cost more than a simple linear search. Given that linear search must be performed regardless of which container we use, an array is the better choice of the two.
You compare items in a linked list by searching for them. Iterate through the list, comparing elements with the search key. If you encounter end-of-list, then the key is not found, otherwise you have found the element desired. Note that this is a half linear search. Statistically, if an element is to be found, it will be found at the halfway point, assuming uniformly random distribution of data. In the worst case, if the element is not found, it will always take a full search to prove that. You could keep the linked list in order, by inserting each element before the element that has higher key value. This would reduce search time to half, because searching would stop with the element with higher key value. Searching is not a very efficient use of a linked-list. It would be better to use some kind of ordered list, perhaps a dynamic array with binary search, or a balanced binary tree, which has similar search performance but one with the most cost to design and implement. Linked-lists are better for keeping elements in the order they were encountered or inserted, such as processing tokens in a compiler. Sorry, but every solution has its tradeoffs.