COMP 14Introduction to Programming

Miguel A. Otaduy

June 4, 2004

Today

• Review arrays

• Searching arrays for a particular value

• Sorting arrays

Exercises

1. Find Sum and Average of Array

2. Determine Largest and SmallestElements in Array

0 1 2 3

Example

double[] sale={3.5, 4.6, 5.2, 3.8};double sum = 0;double average;

Find Sum and Average of Array

0 1 2 3

Example

double[] sale={3.5, 4.6, 5.2, 3.8};double sum = 0;for(int ind = 0; ind < sale.length; ind++){ sum = sum + sale[ind];}double average;if(sale.length != 0) average = sum / sale.length;else average = 0.0;

Find Sum and Average of Array

0 1 2 3

Example

double[] sale={3.5, 4.6, 5.2, 3.8};int maxIndex = 0, minIndex = 0;double largestSale;double smallestSale;

Determining Largest/Smallest Element in Array

0 1 2 3

Example

double[] sale={3.5, 4.6, 5.2, 3.8};int maxIndex = 0, minIndex = 0;for (int ind = 1; ind < sale.length; ind++){ if (sale[ind] > sale[maxIndex]) maxIndex = ind; else if (sale[ind] < sale[minIndex]) minIndex = ind;}double largestSale = sale[maxIndex];double smallestSale = sale[minIndex];

Determining Largest/Smallest Element in Array

0 1 2 3

Searching Arrays

• Find one/several particular element(s) in an array of many elements

• Complexity (How Long To Search?)– find a parking space - linear– look up a word in a dictionary - complex

500K+ words in Oxford Dictionary search - very complexover 3 trillion web pages

Time Complexity

• Important feature in Computer Science research:

– How long does an algorithm take to complete?

– Given an input (e.g. an array) of size n, how long does it take for the algorithm (e.g. search for a particular value) to complete?

• Linear algorithm: time = k1 * n

• Quadratic algorithm: time = k2 * n2

• etc.

Linear Searching

• Algorithm:– Get a test value and a list of values

• list can be ordered or unordered

– loop through the list• repeatedly ask: Is this a match?• quit when the answer is yes (use break stmt)

– if you finish all items, there is no match

• Inefficient– worst time to search is the length of the list

• Relatively easy to program

Example

int[] list={3, 6, 27, 8, 33, 54, 23};int foundAt = -1, element = 33;

Linear Search

0 1 2 3

Example

int[] list={3, 6, 27, 8, 33, 54, 23};int foundAt = -1, element = 33;for(int i=0; i<list.length; i++){

if(list[i] == element){

foundAt = i;break;

}}

Linear Search

0 1 2 3

Array of Objects

• Declare array of Student objects (ref. variables)

• Instantiate array of size 10• Instantiate each of the Student objects

– Ask for age (int) and name (String)– Instantiate object

Example

Student[] students;students = new Student[10];for(int i = 1; i < students.length; i++){

//get int age//get String name

students[i]=new Student(name, age);

}

Array of Student objects

0 1 2 3

Example

int foundAt=-1;String name=“Mark”;for(int i = 1; i < students.length; i++){

if(students[i].getName().equals(name)){

fountAt=i;break;

}}

Search for student Mark

0 1 2 3

Binary Search

• Requires ordered (sorted) list• Set searchRange to the entire list• Repeat:

– pick a “test value” in the middle of searchRange

– if test value == value searching for• Stop!

– if test value > value searching for• searchRange = lower half of searchRange

– if test value < value searching for• searchRange = upper half of searchRange

Example

2 4 5 12 16 19 22 26 29 32 37 41 46 50

Looking for 46 Trial 1

2

3

2 4 5 12 16 19 22 26 29 32 37 41 46 50

2 4 5 12 16 19 22 26 29 32 37 41 46 50

SortingSort students by birthdate• Get a group of 5 students• Another student will sort them by birth

date• Analyze the sorting strategy

– Can we devise an algorithm following that strategy?

• Goal: sort following a methodology– We should be able to write it as an

algorithm and then program it

• Demonstrate selection sort• Write algorithm

Selection SortGeneral Algorithm• Scan the list to find the smallest value• Swap that value with the value in the

first position in the list• Scan rest of list to find the next smallest

value• Swap that value with the value in the

second position in the list• And so on, until you get to the end of

the list

Selection SortMethods• Scan the list to find the smallest value• Swap that value with the value in the

first position in the list• Scan rest of list to find the next smallest

value• Swap that value with the value in the

second position in the list• And so on, until you get to the end of

the listloop

Selection Sort

• Sorts in ascending order

• Can be changed to sort in descending order– look for max instead of min

Homework 6

• Read/write files• Array of ints• Array of Participant objects

• Next:– Sort array of ints– Sort array of participants

Monday

• Java applets• HTML• GUIs

• Reading Assignment:– Chapter 6 (pp. 264-290)– skim Chapter 13 (pp. 725-806)