Question

You are given an array A of integers in sorted order. However, you do not know the length n of the array. Assume that in our programming language arrays are implemented in such a way that you receive an out-of-bounds error message whenever you wish to access an element A[i] with i>n. For simplicity we assume that the error message simply returns the value INT_MAX and that every value in the array is smaller than INT_MAX. (a) Design an algorithm with O(logn) time efficiency that takes an integer key as input and finds a position in the array containing key, if key is an element of A. (b) Explain in detail why your algorithm runs in O(logn).

Answer 1

(a):

Since the given array is sorted, we can make use of binarySearch

Algorithm of binarySearch:

int binarySearch(A[],key):

• start=0
• end=A.length-1
• while(start<end):
  • int middle=(start+end)/2
  • if(A[middle]==key)
    • return middle
  • if(A[middle]>key)
    • start=middle+1
  • else
    • end=middle-1
• return INT_MAX

And time complexity of above code is O(logn) where n is the length of array

(b):

If we observe the code,

initially in step-1, size of search array is n.

In step-2, size of search array is n/2

In step-3, size of search array is n/4

In step-4, size of search array is n/8 and continues till jey value is found or size of array becomes 1 or less.

So number of steps possible are n + n/2 + n/4 + n/8 + ..... + 1 which is logarithm sequence whose value is O(logn). And at the same time in each step we perform operations of time of O(1).

So the final time complexity is O(logn)*O(1) = O(logn)

Mention in comments if any mistakes or errors are found. Thank you.