7When searching a table with length n by binary search, how many comparisons are needed in the worst case?OA nОB lognOC nlognOD logn+18.In BST, the node with the largest key valueO A The left pointer must be nullB The right pointer must be nullOC Both left and right pointers are nullD Both left and right pointers are not null

Answered: 7 When searching a table with length n…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

7
When searching a table with length n by binary search, how many comparisons are needed in the worst case?
OA n
ОB logn
OC nlogn
OD logn+1
8.
In BST, the node with the largest key value
O A The left pointer must be null
B The right pointer must be null
OC Both left and right pointers are null
D Both left and right pointers are not null

Expert Solution

Answer 7

The worst case of a binary search is when you search for the first element or the last element in the array or search for an element which does not exist in the array.
In binary search algorithm, each iteration divide the array into half of the array.

At Iteration 1, Length of array = n
At Iteration 2, Length of array = $\frac{n}{2}$
At Iteration 3, Length of array = ( $\frac{n}{2}$ ) ⁄ 2 = $\frac{n}{2^{2}}$
Therefore, after k Iteration, Length of array = $\frac{n}{2^{k}}$

After k divisions, the length of the array becomes 1 Therefore

$\frac{n}{2^{k}}$ = 1 $\Rightarrow n = 2^{k}$

Taking log₂ both sides, we get:

log₂n = log₂ 2^k $\Rightarrow k = \log_{2} (n)$

Hence, In worst case log₂n comparisons are needed.

Therefore, the correct answer is b. logn

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