fun search(values: IntArray, lookingFor: Int): Boolean {

return false

}

assert search(arrayOf(1, 2, 4), 4)

This lesson covers one of the efficient algorithms enabled by sorting: binary search! This is a fun example of an algorithm that you can approach either recursively or iteratively. We’ll walk through it visually and then let you tackle it on the homework problem.

Binary Search

One of the reasons that sorting is such an important algorithmic primitive is that it enables other efficiency algorithms. For example, we’ve already seen linear array search, which has runtime O(n):

fun search(values: IntArray, lookingFor: Int): Boolean {

for (value in values) {

if (value == lookingFor) {

return true

}

}

return false

}

assert search(new int[]{1, 2, 4}, 4);

Given that O(n) represents searching the entire array, it is clearly the *worst* case runtime for array search.
But, it is also the *best* case *if* we have *no idea* where the item could be!

But what if we knew something about the structure of the data in the array.
Specifically, what if we knew that it was *sorted*?
Could we do better?

Or, as my old graduate school friend David Malan famously explained it:

Binary Search Runtime

Binary search is one recursive algorithm where the O(log n) nature of the algorithm is fairly easy to understand.
Start with an array of N items.
After one round, you have N / 2 left to consider.
Then N / 4.
Then N / 8.
Until you either get to 1, *or* you find the item somewhere along the way.
So in the worst case we do O(log n) steps, and in the best case we may do better.
Cool!

Search Visualizations

As promised, a few cool search visualizations culled from the amazing interwebs. This one has sound!

These are also pretty good, as they show runtime for different inputs as well.

Implementing Binary Search

Now it’s your chance to implement this classic algorithm! There are good approaches that use both recursion and iteration. The iterative approach maintains the start and end index of where the item might be in the array and adjusts these on each iteration. The recursive approach has the base case being an empty or single-item array, and makes the problem smaller by determining whether the item should be in the left half-array or right half-array. Good luck and have fun!

Created By: Geoffrey Challen

/ Version: `2021.4.0`

Let's implement a classic algorithm: binary search on an array.

Implement method named `search`

that takes a `SearchList`

as its first parameter and an `Int`

as its second.
If the `SearchList`

is empty you should throw an `IllegalArgumentException`

.

`SearchList`

is a provided class. It provides a `get(Int)`

method that returns the `Int`

at that index, and
a `size`

method that returns the size of the `SearchList`

. Those are the only two methods you should need!
You can also access the list elements using bracket notation just like a list or array: `list[position]`

.

`search`

returns a `Boolean`

indicating whether the passed `value`

is located in the sorted `SearchList`

.
To search the sorted `SearchList`

efficiently, implement the following algorithm:

- Examine the value in the middle of the current array (index
`(start + end) / 2`

) - If the midpoint value
*is*the value that we are looking for, return`true`

- If the value that we are looking for is
*greater*than the midpoint value, adjust the current array to start at the midpoint - if the value that we are looking for is
*less*than the midpoint value, adjust the current array to end at the midpoint - Continue until you find the value, or until the
`start`

reaches the`end`

, at which point you can give up and return`false`

This is a fun problem! Good luck! Keep in mind that every time you call `SearchList.get`

or use bracket notation
that counts as one access, so you'll need to reduce unnecessary accesses to pass the test suite.
Specifically, you should only call `list.get`

once per iteration, saving the result into a temporary variable.

Need more practice? Head over to the practice page.