7.5 Searching

What does college board want you to know

Differences in searching using arrayList and arrays
Types of searches: sequential (linear) and binary
Searching for a double vs int vs object

Differences Reminder

Answer all the question, they are popcorn hacks

Arrays

Query looks like array[index]
array.length
int[] array = new int[];

arrayList

Query looks like arrayList.get(index)
arrayList.size()
ArrayList<Integer> arrayList = new ArrayList<Integer>();

Question 1

  
// For arrays:
// use square brackets[] to access item
// use .length

if (items[index] == num){
    return index;
}

Question 2

  
// For arraylist:
// use .get() to access item
// use .size

if (items.get(index) == num){
    return index;
}

Searching for a double vs int vs object

Data Type Basics:
- double is used for decimal numbers (like 3.14 or 10.5).
- int is for whole numbers (like 5 or -10).
- Object is a generic type that can hold any kind of data.
Comparing Values:
- With double and int, searching algorithms can directly compare values using simple checks like “Is this number greater than that number?”
- With Object, the comparison might involve more steps, like checking specific properties of the objects.
Performance Considerations:
- double and int use less memory and have simpler comparison logic, which can make searching faster.
- Object might be slower due to the need for more complex comparison logic and potentially larger memory usage.
Handling Null Values:
- With Object, you need to handle cases where the data is null (empty) to avoid errors during searching.
Binary Search Advantage:
- Binary search works best on sorted data, and with double and int, the sorting and comparison are straightforward.
- With Object, sorting and comparison might require more effort, especially for custom data types.

What is Linear Search?

Linear or Sequential Search is a simple searching algorithm that checks each element in a list one by one until it finds a match or reaches the end of the list.

usually used when there is no specific order or structure to the data
O(n), where n is the number of elements in the array

Return the position of key in arr or -1 if key is not in arr.

Return true if key is in arr; otherwise, return false.

  
import java.util.ArrayList;

public class LinearSearch {

    // Iterative implementation for ArrayList<String>
    public static int iterativeLinearSearch(ArrayList<String> list, String target) {
        // Start from the first element of the list
        for (int i = 0; i < list.size(); i++) {

            // Compare each element with the target element until a match is found or the end of the list is reached
            if (list.get(i).equals(target)) {
                return i; // Element found at index i
            }
        }
        return -1; // Element not found
    }

    // Recursive implementation for ArrayList<String>
    private static int search(ArrayList<String> list, String target, int startIndex) {
        // Check if the startIndex is greater than or equal to the size of the list
        if (startIndex >= list.size())
            // If startIndex is out of bounds, return -1 -> target not found
            return -1;
    
        // Check if the string at the startIndex position in the list is equal to the target string
        if (list.get(startIndex).equals(target))
            // If the target string is found at startIndex, return the index
            return startIndex;
    
        // If the target string is not found at startIndex, recursively call the search method
        // with the incremented startIndex to continue searching in the rest of the list
        return search(list, target, startIndex + 1);
    }


    public static void example1(String[] args) {
        ArrayList<String> namesList = new ArrayList<>(Arrays.asList("Grace", "Emma", "Finn", "Theo", "Rachit", "Tanisha", "Vivian", "Aliya", "Justin"));

        int index = LinearSearch.iterativeLinearSearch(namesList, "Emma");
        if (index != -1) {
            System.out.println("Element found at index: " + index);
        } else {
            System.out.println("Element not found");
        }
    }

    public static void example2(String[] args) {
        ArrayList<String> namesList = new ArrayList<>(Arrays.asList("Grace", "Emma", "Finn", "Theo", "Rachit", "Tanisha", "Vivian", "Aliya", "Justin"));

        int index = LinearSearch.recursiveLinearSearch(namesList, "Vivian");
        if (index != -1) {
            System.out.println("Element found at index: " + index);
        } else {
            System.out.println("Element not found");
        }
    }
}

  
// Example with Iterative Implementation
LinearSearch.example1(null);

Element found at index: 1

  
// Example with Recursive Implementation
LinearSearch.example2(null);

Element found at index: 6

Popcorn Hack

Implement linear search for an array list of integers

  
import java.util.ArrayList;

public class LinearPopHack {
    public static Integer Linear(ArrayList<Integer> intList, int findNum){
        for (int i = 0; i < intList.size()-1; i++){
            if (intList.get(i) == findNum){
                return i;
            }
        }
        return -1;
    }

    public static void main(String[] args){
        ArrayList<Integer> arr = new ArrayList<>();

        arr.add(3);
        arr.add(19);
        arr.add(24);
        arr.add(8);
        arr.add(16);

        int foundInd = Linear(arr, 8);
        if (foundInd == -1){
            System.out.print("Element not found.");
        } else {
            System.out.print("Element found at index " + foundInd + ".");
        }
    }
}
LinearPopHack.main(null);

Element found at index 3.

When is it preferred to use linear search over binary search?
- When the list is not sorted (binary search can’t be used on unsorted lists)
- When the desired element to be found is the first index of the list (will be same time as binary search if the desired element is at the middle, but linear search is easier to implement)

Recursive algorithms

a method that calls upon itself
has a “base case” that is a conditional statement that skips the recursive call so algorithm stops at certain point

useful when solving problems that can be broken down into smaller, repetitive problems

Popcorn Hack

What are some examples of algorithms that would be useful to have recursion?

Merge sort
Quick sort
Fibonacci sequences

Binary Search

finds the index of an element of a SORTED array
why use it? can be faster than linear search which has O(n) complexity while binary has O(log N)
how it works simply: (Divide and Conquer method 💪)

Start in the middle –> see if that number is lower or higher than the desired number
a. if lower than disregard upper half section and only look at lower section
b. if higher than disregard lower half and only look at upper section
Find middle of that lower/higher section (those sections don’t include the original middle number)
a. if even number of numbers in that section, you can specify in algorithm whether want to use lower or higher number
Keep repeating process until find number
a. if desired number trying to find is not in the list → high and low are swapped incorrectly → return -1

Simple example

  
public class BinarySearch {
    static char[] arr = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'};

    public static String findMe(char target, int start, int end) {
        if (start > end) {
            return "Not Found";
        }

        // find middle number - java integer division automatically truncates
        int middle = (start + end) / 2;

        if (arr[middle] == target) {
            return "Found it at index " + middle;
        }

        // recursion spotted - search lower section
        if (arr[middle] > target) {
            return findMe(target, start, middle - 1);
        } 

        // recursion spotted part 2 - search higher section
        if (arr[middle] < target) {
            return findMe(target, middle + 1, end);
        }
        
        return "Not Found";
    }

    public static void main(String[] args) {
        char target = 'f';
        int start = 0;
        int end = arr.length - 1;
        System.out.println(findMe(target, start, end));
    }
}
BinarySearch.main(null);

Found it at index 5

Popcorn Hack

What iteration did it find f?

2nd iteration.

Hashmap searching

Introduction to HashMap:

HashMap is a data structure that stores key-value pairs.
Keys are hashed to determine their storage location in the map.
Java’s HashMap provides O(1) time complexity for get() and put() operations.
No keys can be the same, or else the old data is lost, and is replaced by the new one

  
import java.util.HashMap;

public class HashMapSearching {
    public static void main(String[] args) {
        // Create a HashMap of students and their scores
        // Declaring the HashMap with <String, Integer> type
        HashMap<String, Integer> scores = new HashMap<>();
        scores.put("Alice", 85);
        scores.put("Bob", 90);
        scores.put("Charlie", 95);
        scores.put("Alice", 80);

        // Search for a student
        String name = "Alice";

        // containsKey() method is used to check if the key is present in the HashMap
        if (scores.containsKey(name)) {
            int score = scores.get(name);
            System.out.println(name + "'s score is: " + score);
        } else {
            System.out.println(name + " not found in the records.");
        }
    }
}

HashMapSearching.main(null);

Alice's score is: 80

Hashmaps with objects

What is an Abstract Class Review
- Abstract classes can’t be made into objects
- They enforce a structure for the subclasses generated by it

Hacks for this part

Create a method to delete data based off the key

  
import java.util.HashMap;

public abstract class Collectable implements Comparable <Collectable> {
	public final String masterType = "Collectable";
	private String type;	// extender should define their data type

	// enumerated interface
	public interface KeyTypes {
		String name();
	}
    
	protected abstract KeyTypes getKey();  	// this method helps force usage of KeyTypes

	// getter
	public String getMasterType() {
		return masterType;
	}

	// getter
	public String getType() {
		return type;
	}

	// setter
	public void setType(String type) {
		this.type = type;
	}
	
	// this method is used to establish key order
	public abstract String toString();

	// this method is used to compare toString of objects
	public int compareTo(Collectable obj) {
		return this.toString().compareTo(obj.toString());
	}

	// static print method used by extended classes
	public static void print(Collectable[] objs) {
		// print 'Object' properties
		System.out.println(objs.getClass() + " " + objs.length);

		// print 'Collectable' properties
		if (objs.length > 0) {
			Collectable obj = objs[0];	// Look at properties of 1st element
			System.out.println(
					obj.getMasterType() + ": " + 
					obj.getType() +
					" listed by " +
					obj.getKey());
		}

		// print "Collectable: Objects'
		for(Object o : objs)	// observe that type is Opaque
			System.out.println(o);

		System.out.println();
	}
}

public class Car extends Collectable {
    private String make;
    private String model;
    private int year;

    public Car(String make, String model, int year) {
        this.make = make;
        this.model = model;
        this.year = year;
    }

	@Override
    protected KeyTypes getKey() {
        return null; 
    }

    public String getMake() {
        return make;
    }

    public String getModel() {
        return model;
    }

    public int getYear() {
        return year;
    }

    public String toString() {
        return year + " " + make + " " + model;
    }
}

public class Garage {
    private static HashMap<String, Car> garage = new HashMap<>();

    public Garage() {
        garage.put("Lambo", new Car("Lamborghini", "Aventador", 2021));
        garage.put("Ferrari", new Car("Ferrari", "F8 Tributo", 2021));
        garage.put("Porsche", new Car("Porsche", "911 Turbo S", 2021));
        garage.put("McLaren", new Car("McLaren", "720S", 2021));
    }

    //print what's in my garage
    public void printGarage() {
        for (String key : garage.keySet()) {
            System.out.println(key + ": " + garage.get(key));
        }
    }

    // removing car hack!!!
    public void removeCar(String key) {
        Car removedCar = garage.remove(key);
        if (removedCar == null) {
            System.out.println(key + " not found in the garage.");
        } else {
            System.out.println("Removed car: " + removedCar);
        }
    }

    public static void main(String[] args) {
        Garage myGarage = new Garage();
        myGarage.printGarage();

        // Removing a car from the garage tester code
        // String key = "Lambo";
        // Car car = garage.remove(key);
        // if (car == null) {
        //     System.out.println(key + " not found");
        // } else {
        //     System.out.println("Removed: " + key + ", " + car);
        // }
        String key = "Lambo";
        myGarage.removeCar(key);
    }
}

Garage.main(null);

Ferrari: 2021 Ferrari F8 Tributo
Porsche: 2021 Porsche 911 Turbo S
Lambo: 2021 Lamborghini Aventador
McLaren: 2021 McLaren 720S
Removed car: 2021 Lamborghini Aventador

HACKS (you should be able to do with chatgpt)

Is sequential/linear or binary more efficient? Why?
- Depends, but usually binary is more efficient, especially for large SORTED data sets. Sequential searches EVERY single element in the array, and worst case it is O(n) which is the number of elements in the array. However, binary splits half in each time, so worst case is O(log n) which is a lot faster for large sorted data sets.
Why might you not always be able to use binary search?
- If the numbers aren’t sorted in order from least to greatest
- Can’t be used for unordered data structures like hashmaps or unordered maps
- Hard to sort if array is frequently modified
Which of the following implements a method named contains for searching an array sequentially, confirming whether or not the array contains a requested element?
- The fourth one is correct: It returns true immediate when it finds the desired element and exits out of the loop, if it doesn’t find desired element after searching entire arr it will return false. This is correct

Answer the comment in the code

  
public static int foo(int[] arr, int x) {

    for(int i = 0; i < arr.length; i++) {

        if(arr[i] == x) {

            return i;

        }

    }

    return -1;

}
// Given the method defined above, how many times is the word "Indubitably!" output by the code below?

int[] vals = {1,4,51,3,14,91,130,14};

for(int i = 0; i < 20; i++) {

    if(foo(vals,i%4) < 0) {

        System.out.println("Indubitably!");

    }
}

Answer:

10 times. The (i%4) is the remainder, which will always be 0, 1, 2, or 3. 1 and 3 are the only ones that exist in the array, so Indubitably will print when the remainder is 1 or 3. This will be when i=1,3,5,7,9,11,13,15,17,19, so Indubitably will print 10 times.

Workshop 3 - Searching

7.5 Searching

What does college board want you to know

Differences Reminder

Arrays

arrayList

Question 1

Question 2

Searching for a double vs int vs object

What is Linear Search?

Popcorn Hack

Recursive algorithms

Popcorn Hack

Binary Search

Simple example

Popcorn Hack

Hashmap searching

Introduction to HashMap:

Hashmaps with objects

Hacks for this part

HACKS (you should be able to do with chatgpt)

Answer the comment in the code

Answer:

Further Reading

Algorithmic Sort

Workshop 3 - Arrays

Workshop 3 - ArrayList