If this is a homework related question, you really should consider trying to solve it yourself before looking at this answer. Otherwise, the value of the lesson, and the reinforcement provided by the assignment, will be lost to you.
In a sequential search, where the elements are in a uniformly random distribution, the average number of comparisions to find a particular element is one half of the number of elements.
Stated another way... In a sequential search, where the elements are in an arbitrary distribution, the average number of comparisions to find a random element is one half of the number of elements.
N/2
If you're strictly using a sequential search, then the order of the array's content will make no difference. Whether it's in low-high order, high-low order, or randomized, the time complexity for a sequential search will remain O(n).
Linear search takes linear time with a worst case of O(n) for n items, and an average of O(n/2). Binary search takes logarithmic time, with a worst and average case of O(n log n). Binary search is therefore faster on average.
n comparisons.
Sequential search is the only way to search an unsorted array unless you resort to a multi-threaded parallel search where all threads concurrently search a portion of the array sequentially.
O(N) where N is the number of elements in the array you are searching.So it has linear complexity.
Comparisons focus on highlighting similarities between two or more things, while contrasts emphasize differences between them. Comparisons typically examine how things are alike, while contrasts explore how they are different.
Sequential search of an object with in an array of objects is called as linear search.
As I know the search method depends on your(programmer's) logic. In sequential search it will be better to stop the search as soon as search value encounters or if search value is not in the array then it should stop at the end.
There is no such thing as an insertion search. There is only insertion sort, which is a method of sorting an unsorted list. Sequential search (or linear search) is only used with unsorted lists. If the list is sorted, a logarithmic search is quicker, by starting from the middle. If the items is not here, it must be in the lower half or the upper half, thus one half of the list can be discarded. You then repeat by starting in the middle of the remaining half. Thus for a list of 15 items, you end up with a list of 7, then 3, then 1, then 0. Thus it takes 5 comparisons to determine that an item does not exist. With linear search it would take 15 comparisons to determine that an item does not exist. Thus logarithmic search is quicker, but only works with sorted lists.
A binary search is much faster.
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.