Question

John needs a data structure for searching 10240 keys, but he knows that 80% of these searches for keys that are actually present involve only 20% of the keys. He decides to separate these 2048 keys into one ordered table, and keep the other 8192 keys in a separate but also ordered table. To find a key he will look in the small table first, using binary search, and if the key is not found we look in the large table using the same algorithm.

Note: [ ] represent ceiling function notation

Problem 1: In the worst case, John's algorithm takes time ______ while binary search on a single table of size 10,240 takes time _______; pick the most accurate representation:

A. [log₂(2048)] + [log₂ (8192)] ; [log₂ (8192)]

B. [log₂(2048)] + [log₂ (8192)] ; [log₂ (10240)]

C. [log₂(2048)] + 0.2 * [log₂ (8192)] ; [log₂ (8192)]

D.  [log₂(2048)] + 0.2 * [log₂ (8192)] ; [log₂ (10240)]

E.  [log₂(2048)] + 0.2 * [log₂ (8192)] ; 0.8 * [log₂ (10240)]

Problem 2: In the expected case for successful searches, John's algorithm takes time ________ while binary search on a single table of size 10240 takes time ______; pick the most accurate expression.

A. [log₂(2048)] + [log₂ (8192)] ; [log₂ (8192)]

B. [log₂(2048)] + [log₂ (8192)] ; [log₂ (10240)]

C. [log₂(2048)] + 0.2 * [log₂ (8192)] ; [log₂ (8192)]

D.  [log₂(2048)] + 0.2 * [log₂ (8192)] ; [log₂ (10240)]

E.  [log₂(2048)] + 0.2 * [log₂ (8192)] ; 0.8 * [log₂ (10240)]

Answer 1

Binary Search:- In Binary Search we can search elements in sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. Go to the left half if the value of the search key is less than the item in the middle of the interval. Otherwise go for the right half. Repeatedly check until the value is found or the interval is empty.

So we have to   divide by 2 until we have only one element left that is -

n/2^k=1( if array is having n elements)

that is 2^k=n

k=log₂n This is the worst case time complexity of the binary search Therefore for worst case as john is also using binary search on two arrays for problem 1 option B is correct.

When we think of the Best case of Binary search It occurs when the element you are searching for is the middle element of the list/array . In that case you will get the desired result in a single search.

Time Complexity of the Algorithm in such a case will be O(1).

But in problem 2 you stated expected case for successful searches that means we should think for average case. and for average case time complexity of binary search is also log₂n.

Therefore according to me for problem 2 also option B is correct.