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Published at Aug 23 2019
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Instructions

Test suite

Solution

Implement a binary search algorithm.

Searching a sorted collection is a common task. A dictionary is a sorted list of word definitions. Given a word, one can find its definition. A telephone book is a sorted list of people's names, addresses, and telephone numbers. Knowing someone's name allows one to quickly find their telephone number and address.

If the list to be searched contains more than a few items (a dozen, say) a binary search will require far fewer comparisons than a linear search, but it imposes the requirement that the list be sorted.

In computer science, a binary search or half-interval search algorithm finds the position of a specified input value (the search "key") within an array sorted by key value.

In each step, the algorithm compares the search key value with the key value of the middle element of the array.

If the keys match, then a matching element has been found and its index, or position, is returned.

Otherwise, if the search key is less than the middle element's key, then the algorithm repeats its action on the sub-array to the left of the middle element or, if the search key is greater, on the sub-array to the right.

If the remaining array to be searched is empty, then the key cannot be found in the array and a special "not found" indication is returned.

A binary search halves the number of items to check with each iteration, so locating an item (or determining its absence) takes logarithmic time. A binary search is a dichotomic divide and conquer search algorithm.

The Scala exercises assume an SBT project scheme. The exercise solution source should be placed within the exercise directory/src/main/scala. The exercise unit tests can be found within the exercise directory/src/test/scala.

To run the tests simply run the command `sbt test`

in the exercise directory.

For more detailed info about the Scala track see the help page.

Wikipedia http://en.wikipedia.org/wiki/Binary_search_algorithm

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

```
import org.scalatest.{Matchers, FunSuite}
/** @version 1.3.0 */
class BinarySearchTest extends FunSuite with Matchers {
test("finds a value in an array with one element") {
BinarySearch.find(List(6), 6) should be(Some(0))
}
test("finds a value in the middle of an array") {
pending
BinarySearch.find(List(1, 3, 4, 6, 8, 9, 11), 6) should be(Some(3))
}
test("finds a value at the beginning of an array") {
pending
BinarySearch.find(List(1, 3, 4, 6, 8, 9, 11), 1) should be(Some(0))
}
test("finds a value at the end of an array") {
pending
BinarySearch.find(List(1, 3, 4, 6, 8, 9, 11), 11) should be(Some(6))
}
test("finds a value in an array of odd length") {
pending
BinarySearch.find(List(1, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 634),
144) should be(Some(9))
}
test("finds a value in an array of even length") {
pending
BinarySearch.find(List(1, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377), 21) should be(
Some(5))
}
test("identifies that a value is not included in the array") {
pending
BinarySearch.find(List(1, 3, 4, 6, 8, 9, 11), 7) should be(None)
}
test("a value smaller than the array's smallest value is not found") {
pending
BinarySearch.find(List(1, 3, 4, 6, 8, 9, 11), 0) should be(None)
}
test("a value larger than the array's largest value is not found") {
pending
BinarySearch.find(List(1, 3, 4, 6, 8, 9, 11), 13) should be(None)
}
test("nothing is found in an empty array") {
pending
BinarySearch.find(List(), 1) should be(None)
}
test("nothing is found when the left and right bounds cross") {
pending
BinarySearch.find(List(1, 2), 0) should be(None)
}
}
```

```
import scala.annotation.tailrec
object BinarySearch {
def find(list: List[Int], target: Int): Option[Int] = {
@tailrec
def binaryFind(left: Int, mid: Int, right: Int): Option[Int] = list match {
case Nil => None
case x => (left, mid, right) match {
case (_, m, _) if x(m) == target => Some(m)
case (l, _, r) if l >= r => None
case (l, m, r) if x(m) < target => binaryFind(m + 1, (m + 1 + r) / 2, r)
case (l, m, r) if x(m) > target => binaryFind(l, (r - m - 1) / 2, m - 1)
}
}
binaryFind(0, list.length / 2, list.length - 1)
}
}
```

A huge amount can be learned from reading other peopleâ€™s code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

- What compromises have been made?
- Are there new concepts here that you could read more about to improve your understanding?

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