(a):
Since the given array is sorted, we can make use of binarySearch
Algorithm of binarySearch:
int binarySearch(A[],key):
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.
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