O（1）和O（n）之间的线性搜索差异

Question

I am confused with the following two cases in linear search. Let me write the linear search algorithm here first.

Pseudo code for linear search

input: unsorted array A & key

output: index i such that A[i] = key

LinearSearch(A,low,high,key) 
if high < low    
   return NOT_Found
if A[low] = key
   return low
return LinearSearch(A,low+1,high,key)

The complexity is O(n) in most of the book. What is the difference between

1. Best-case runtime for linear search if key is present in the array [Ans: O(1)]
2. Best-case runtime for linear search if key isn't present in the array [Ans: O(n)]

What is the reason that when the key is present in the array, the complexity will reduce to O(1)? Thank you very much!

Answer 1

Your question is based on the best case. The best case would be that your key is on the first position (O(1)). More practical would be the average case which is in both cases O(n)

Answer 2

What is the reason that when the key is present in the array, the complexity will reduce to O(1)?

The presence of the key in the array makes a difference since consider the two cases:

• The key exists in the array : then your running time in best case is O(1) , which means that you will have to move constant number of times in order to find it (eg it will be the 2nd element, the 10th element, the 1st element generally sth constant). If the key exists then the worst case would be that it is in O(n) position in the array so you need at least Ω(n) steps and of course maximum steps is to search all n positions in array so O(n) which proves worst case Θ(n) .

• The key doesn't not exist : In this case given an array and a key that does not exist in the array your linear search algorithm attempting to find it he will have to check all elements in order to find it because the algorithm has two termination conditions: the key is found or the array has ended. Of courses the key in this case will not be found and the algorithm will terminate after checking all elements. An observation here is that if the key does not exist the algorithm will do exactly O(n) steps both in best or worst case, if the key does not exist best and worst cases are the same since in both you search whole array!!

In both cases as explained the worst case is θ(n) .