Question #$5.$ By now you should be convinced that binary search requires fewer comparisons than linear search in both the worst case and the average case. However, keep in mind that binary search assumes that the elements are already in sorted order; if they$$re not sorted, then it would take time to sort them before applying binary search, and sorting is a slow operation.

Whereas the number of operations in the worst case to conduct binary search on a collection of N items is approximately

log

$2$

$$

log

$2$

$$

N$,$ the number of operations required to sort N items is

$$

log

$2$

$$

Nlog

$2$

$$

N$,$ i$.$e$.,$ the value of N times its base$-2$ logarithm.

Assume you have an unsorted collection of N $=15$ items, and you know you will need to make at least $10$ searches and are concerned about the worst case. How would you determine whether it would be better to conduct $10$ linear searches, or if it would be better to first sort the elements $($which you only need to do once$)$ and then conduct $10$ binary searches?

In making this decision, follow these steps:

First, calculate the number of operations it would require to conduct $10$ linear searches in the worst case.

Then, calculate the number of operations it would require to sort the elements once; for simplicity, round

log

$2$

$15$

log

$2$

$$

$15$ up to $4.$

Last, calculate the number of operations it would require to conduct $10$ binary searches in the worst case.

Use the results of these three calculations to determine the answer.

Show the results of your calculations for steps $1-3$ and then explain your answer in step $4.$