1. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 1 Chapter 11 Searching and Sorting

2. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 2 Searching When we maintain a collection of data, we often need to quickly locate data Here’s the problem statement: Given a value X, return the index of X in the array, if such X exists. Otherwise, return NOT_FOUND (-1). We will look at two different algorithms. We will count the number of comparisons the algorithms make to analyze their performance. –The ideal searching algorithm will make the least possible number of comparisons to locate the desired data. –Two separate performance analyses are normally done, one for successful search and another for unsuccessful search.

3. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 3 Linear Search Search the array from the first to the last position in linear progression. public int linearSearch ( int[] number, int searchValue ) { int loc = 0; while (loc < number.length && number[loc] != searchValue) { loc++; } if (loc == number.length) { //Not found return NOT_FOUND; } else { return loc; //Found, return the position }

4. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 4 Linear Search Performance Count how many times the search value is compared against the array elements. –Analyze the successful and unsuccessful searches separately. Successful Search –Best Case = 1 comparison –Worst Case = N comparisons (N = array size)‏ Unsuccessful Search –Best Case = Worst Case = N comparisons

5. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 5 Search Result number 231759012 4438 012345678 8477 Unsuccessful Search: Successful Search: NOT_FOUND search( 45 )‏ search( 12 )‏ 4

6. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 6 Binary Search If an array is sorted, then we can search faster using a binary search. The basic idea is straightforward. –First check the value in the middle position. –If X is less than this value, then search the left half next. –If X is greater than this value, then search the right half next. –Repeat the process until X is found or obviously not there. number 512172338 4477 012345678 8490

7. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 7 Sequence of Successful Search - 1 512172338 4477 012345678 8490 search( 44 )‏ lowhighmid 0808 #1 high low 38 < 44 low = mid+1 = 5 mid 4

8. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 8 Binary Search Routine public int binarySearch (int[] number, int searchValue) { int low = 0, high = number.length - 1, mid= (low + high) / 2; while (low <= high && number[mid] != searchValue) { if (number[mid] < searchValue) { low = mid + 1; } else { //number[mid] > searchValue high = mid - 1; } mid = (low + high) / 2; //integer division will truncate } if (low > high) { mid = NOT_FOUND; } return mid; }

9. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 9 Binary Search Performance The section of an array to search is cut into half after every comparison –After K comparisons, there will be N/2 K elements in the list. –To get max number of comparisons, solve for K when N/2 K = 1 K = log 2 N. Successful Search –Best Case = 1 comparison –Worst Case = log 2 N comparisons Unsuccessful Search –Best Case = Worst Case = log 2 N comparisons

10. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 10 Comparing N and log 2 N Performance Array SizeLinear O(N)Binary O(log 2 N)‏

11. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 11 Sorting Often convenient to maintain a collection of data in sorted order –Data needs to be sorted to do a binary search –Example: arranging Person objects in order by age. Here’s the problem statement: Given an array of N integer values, arrange the values into ascending order. We will count the number of comparisons the algorithms make to analyze their performance. –The ideal sorting algorithm will make the least possible number of comparisons to arrange data in a designated order. We will compare different sorting algorithms by analyzing their worst-case performance.

12. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 12 Selection Sort 1.Find the smallest element in the list. 2.Exchange the element in the first position and the smallest element. Now the smallest element is in the first position. 3.Repeat Step 1 and 2 with the list having one less element (i.e., the smallest element is discarded from further processing). 012345678 23175901244388477 min first exchange 012345678 51723901244388477 sorted unsorted This is the result of one pass.

13. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 13 Selection Sort Passes 012345678 51723901244388477 sorted 51223901744388477 2 51217902344388477 3 51217233844778490 7 51217233844778490 8 1 Pass # Result AFTER one pass is completed.

14. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 14 Selection Sort Routine public void selectionSort(int[] number) { int startIndex, minIndex, length, temp; length = number.length; for (startIndex = 0; startIndex <= length-2; startIndex++){ //each iteration of the for loop is one pass minIndex = startIndex; //find the smallest in this pass at position minIndex for (int i = startIndex+1; i <= length-1; i++) { if (number[i] < number[minIndex]) minIndex = i; } //exchange number[startIndex] and number[minIndex] temp= number[startIndex]; number[startIndex] = number[minIndex]; number[minIndex] = temp; }

15. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 15 Selection Sort Performance We derive the total number of comparisons by counting the number of times the inner loop is executed. For each execution of the outer loop, the inner loop is executed length – start times. The variable length is the size of the array. Replacing length with N, the array size, the sum is derived as…

16. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 16 Insertion Sort Exercise 5, page 660 Think of an array of numbers as consisting of a sorted part at the beginning and an unsorted part at the end –Initially, the sorted part is the first element –The unsorted part is the rest of the array While there are elements in the unsorted part –Take the first element of the unsorted part and insert it into the proper position in the sorted part

17. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 17 Bubble Sort With the selection sort, we make one exchange at the end of one pass. The bubble sort improves the performance by making more than one exchange during its pass. By making multiple exchanges, we will be able to move more elements toward their correct positions using the same number of comparisons as the selection sort makes. The key idea of the bubble sort is to make pairwise comparisons and exchange the positions of the pair if they are out of order.

18. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 18 One Pass of Bubble Sort The largest value 90 is at the end of the list. 012345678 23175901244388477 17235901244388477175239012443884771752312904438847717523124490388477175231244389084771752312443884907717523124438847790 exchange ok exchange

19. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 19 Bubble Sort Routine public void bubbleSort(int[] number) { int temp, bottom, i; boolean exchanged = true; bottom = number.length - 2; while (exchanged) { exchanged = false; for (i = 0; i <= bottom; i++) { if (number[i] > number[i+1]) { temp = number[i]; //exchange number[i] = number[i+1]; number[i+1] = temp; exchanged = true; //exchange is made } bottom--; }

20. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 20 Bubble Sort Performance In the worst case, the outer while loop is executed N-1 times for carrying out N-1 passes. For each execution of the outer loop, the inner loop is executed bottom+1 times. The number of comparisons in each successive pass is N-1, N-2, …, 1. Summing these will result in the total number of comparisons. So the performances of the bubble sort and the selection sort are approximately equivalent. However, on the average, the bubble sort performs much better than the selection sort because it can finish the sorting without doing all N-1 passes.

21. ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 th Ed Chapter 11 - 21 Where to find a sort method You don't have to write a sort method every time you need to sort an array The Arrays class from the java.util package has search and sort methods –binarySearch –sort (uses a faster sort method than the ones we looked at)‏ These methods are overloaded to work on arrays of any primitive type as well as arrays of objects

22. Searching Objects Similar to searching primitive arrays except you have several choices –Search for a particular object (by reference) in the array Compare using == –Search for a single object with a particular property or group of properties Match all the properties of a particular object using equals method Use accessor methods to get appropriate data for comparison –Search for all objects that meet search criterion Return an array (or collection) instead of a single object

23. Sorting Objects In order to sort objects of a particular type, we need a way to compare them –The relational operators ( = >) do not work with objects Two approaches to object comparison –Comparable interface The object class must implement the Comparable interface Class must define the method int compareTo( o) –Comparator interface Create a new class which implements the Comparator interface Need to define two methods int compare( o1, o2) boolean equals( Object o) // compare two Comparators

24. Comparable Interface Any class that implements Comparable must contain the method int compareTo( o)‏ Since it is inside the class, compareTo has access to all instance variables The Arrays class has a sort method that sorts Comparable objects –public static void sort (Comparable[] a)‏

25. Semantics of compareTo o1.compareTo( o2) returns –negative number if o1 comes before o2 –zero if o1 and o2 are the same –positive number if o1 comes after o2 For Strings –"abc".compareTo("bcd") is negative –"abc".compareTo("ABC") is positive –"abc".compareTo("abc") is 0

26. Comparator Interface Create a separate class whose sole purpose is to provide a comparison method int compare( o1, o2) Since it is a separate class, compareTo has to use accessor methods to get instance data The Arrays class has a sort method that sorts Comparable objects public static void sort (T[] a, Comparator c)‏

27. Semantics of compare ComparatorClass.compare ( o1, o2) returns –negative number if o1 comes before o2 –zero if o1 and o2 are the same –positive number if o1 comes after o2

28. Comparing Person Objects We define the Person class so we can compare Person objects by name or by age

29. Comparing Person Objects First, we need to determine how to compare Person objects The following does not make sense: Person p1 = …; Person p2 = …; if ( p1 < p2 ) { … }

30. Choosing an attribute to compare Modify the Person class to include a static variable that tells which attribute to compare. class Person implements Comparable { … public static final int NAME = 0; public static final int AGE = 1; public static int compareAttribute = NAME; … public static void setCompareAttribute(int attribute) { compareAttribute = attribute; public int compareTo( Object person){ …} } … }

31. The compareTo Method To compare Person objects, first set the comparison attribute and then call the compareTo Method. Person.setCompareAttribute (Person.NAME); int compResult = p1.compareTo(p2); if (compResult < 0) { //p1 “less than” p2 } else if (compResult == 0) { //p1 “equals” pw } else { //p2 “greater than” p2 } public int compareTo(Person p) { int compResult; if ( comparisonAttribute == AGE ) { int p2age = p.getAge(); if ( this.age < p2age ) { compResult = LESS; } … } else { //compare Name String p2Name = p.getName(); compResult = this.name. compareTo(p2Name); } return compResult; }

32. Using Comparators Implement the Comparator interface. We can define the implementation class so we can compare Person objects using any combination of their attributes –For example, we can define a comparator that compares Person objects by using both name and age

33. The AgeComparator Class This class compares the age attributes of Person objects public class AgeComparator implements Comparator { private final int LESS = -1; private final int EQUAL = 0; private final int MORE = 1; public int compare(Object p1, Object p2) { int comparisonResult; int p1age = ((Person)p1).getAge( ); int p2age = ((Person)p2).getAge( ); if (p1age < p2age) { comparisonResult = LESS; } else if (p1age == p2age) { comparisonResult = EQUAL; } else { assert p1age > p2age; comparisonResult = MORE; } return comparisonResult; }

34. The NameComparator Class This class compares the age attributes of Person objects Because the name attribute is a string we can use the compareTo method of the String class public class NameComparator implements Comparator { public int compare(Object p1, Object p2) { String p1name = ((Person)p1).getName( ); String p2name = ((Person)p2).getName( ); return p1name.compareTo(p2name); }

35. The NameComparator Class This class compares the age attributes of Person objects Because the name attribute is a string we can use the compareTo method of the String class public class NameComparator implements Comparator { public int compare(Person p1, Person p2) { String p1name = p1.getName( ); String p2name = p2.getName( ); return p1name.compareTo(p2name); }

36. The sort Method We use the sort method of the java.util.Arrays class public Person[ ] sort ( int attribute ) { if (!(attribute == Person.NAME || attribute == Person.AGE) ) { throw new IllegalArgumentException( ); } Person[ ] sortedList = new Person[ count ]; //copy references to sortedList for (int i = 0; i < count; i++) { sortedList[i] = entry[i]; } Arrays.sort(sortedList, getComparator(attribute)); return sortedList; }