# Lab 6: Smaller/Bigger Sort

## Learning Outcomes

• Use the compareTo() method from the Comparable interface to determine which of two objects is bigger
• Understand and apply recusion in algorithm development

## Overview

Video Introduction

In this assignment you will implement a recursive algorithm to sort elements of a list. The basic idea is to look at the first element in the list - let's call it first - and rearrange the list such that all of the elements smaller than first appear before first and all elements bigger than first appear after first. Elements equal to first can appear on either side of first. For example,

12345678
838512181813

becomes

01235678
138581281813

We now have a sublist of elements from index 0 to (but not including) index 4 with elements no bigger than 8 and a sublist of elements no smaller than 8 from index 5 to index 9. The elements in the sublists can be in any order.

## Details

You will create a class, SmallerBiggerSort with three class methods:

• public static <T extends Comparable<T>> int smallerBigger(List<T> list, int startInclusive, int endExclusive) — used to get a feel for how the non-recursive portion of the next method works. The method returns the index where the first element ended up being placed.
• public static <T extends Comparable<T>> void sort(List<T> list, int startInclusive, int endExclusive) — a recursive sort method that works by creating sublists of all the elements smaller than a particular value and larger than the value (details below).
• public static <T extends Comparable<T>> void sort(List<T> list) — calls the recursive sort() method with the bounds of the list.
Tips:
• A significant portion of this assignment is related to your analysis (see end of assignment). Make sure you complete the coding part of this assignment well in advance so that you have time to give appropriate attention to the analysis.
• Writing your tests before or as you implement your solution may help you understand the requirements of the code and design better solutions.
• The Comparable interface requires the compareTo() method to be implemented. This method specifies how to compare two objects.

Note: in the examples below are just examples, the order of the elements in the sublist does not need to match what is shown in these examples.

Implement, smallerBigger(List<T> list, int startInclusive, int endExclusive), that places all of the elements smaller than first = list.get(startInclusive) between position startInclusive and the location first and all elements larger than first between the location of first and position endExclusive. This method must be $$O(n)$$ when an ArrayList is used. For example, smallerBigger(list, 2, 6); where list is:

01345🛑78
838512181813

the list becomes

01345🛑78
831581281813

Next implement sort(List<T> list, int startInclusive, int endExclusive) which sorts the list by recursively creating sublists of smaller and larger elements. For example calling sort(list, 0, 9) with list

12345678🛑
838512181813

the list becomes

1235678🛑
138588121813

We now have two sublists that need to be sorted, so we need to call sort(list, 0, 4) and sort(list, 5, 9). sort(list, 0, 4) looks like:

123🛑5678
138588121813

producing

123🛑5678
138588121813

We now have just one sublist that needs to be processed: sort(list, 1, 4) to get:

023🛑5678
138588121813

which leads to sort(list, 2, 4) and this result:

013🛑5678
135888121813

Now the sort(list, 0, 4) call is complete and we can progress with the sort(list, 5, 9) call:

01234678🛑
135888121813

resulting in no changes, but a recursive call to sort(list, 6, 9) which, likewise causes no changes. The next recursive call is sort(list, 7, 9) which swaps the 18 and 13 producing a sort list:

012345678
135888121318
Tips:
• There are several ways that you could implement smallerBigger() in O(n) time.
• One simple way would be to make multiple passes through the list.
• On the first pass, count how many elements are smaller than the first element. Once you know this, you can determine the final resting place for the first element.
• Create a new list and put all of elements that are less than the first element in the beginning of the list, place the first element next, and then place all of the remaining elements into the list.

## Testing

Write thorough JUnit tests for your implementation.