JavaIm not sure how to implement this code can someone help me .pdf

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Java I\'m not sure how to implement this code can someone help me / guide me? thank you (Updated with PrefixMap interface included at the bottom of the code). import java.util.List; import java.util.Map; public class Storage implements PrefixMap { private Map keys; private Map> prefixes; public Storage() { // TODO Initialise your maps } /* * How many keys are stored in the map */ @Override public int size() { // TODO Implement this, then remove this comment return 0; } @Override public boolean isEmpty() { // TODO Implement this, then remove this comment return false; } /* * Get the value corresponding to the key (or null if the key is not found) * if the key contains any character other than A,C,G,T, throw MalformedKeyException * if the key is null, throw IllegalArgumentException */ @Override public String get(String key) { // TODO Implement this, then remove this comment return null; } /* * Insert the value into the data structure, using the given key. If the key * already existed, replace and return the old value (otherwise return null) * if the key contains any character other than A,C,G,T, throw MalformedKeyException * if the key or value is null, throw IllegalArgumentException */ @Override public String put(String key, String value) { // TODO Implement this, then remove this comment return null; } /* * Remove the value corresponding to the given key from the data structure, * if it exists. Return the old value, or null if no value was found. * if the key contains any character other than A,C,G,T, throw MalformedKeyException * if the key is null, throw IllegalArgumentException */ @Override public String remove(String key) { // TODO Implement this, then remove this comment return null; } /* * return the number of keys which start with the given prefix if the prefix * contains any character other than A,C,G,T, throw MalformedKeyException * if the prefix is null, throw IllegalArgumentException */ @Override public int countKeysMatchingPrefix(String prefix) { // TODO Implement this, then remove this comment return 0; } /* * return the collection of keys which start with the given prefix if the * prefix contains any character other than A,C,G,T, throw MalformedKeyException * if the prefix is null, throw IllegalArgumentException */ @Override public List getKeysMatchingPrefix(String prefix) { // TODO Implement this, then remove this comment return null; } /* * Return the number of unique prefixes * e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 8 * because the possible prefixes are G, GA, GAT, GATT, GATTC, GATTA, GATTAC, GATTACA */ @Override public int countPrefixes() { // TODO Implement this, then remove this comment return 0; } /* * Return the sum of the lengths of all keys * e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 15 */ @Override public int sumKeyLengths() { // TODO Implement this, then remove this comment return 0; } } import java.util.List; public interface PrefixMap { public boole.

Education

// TODO Implement this, then remove this comment
return null;
}
/*
* Insert the value into the data structure, using the given key. If the key
* already existed, replace and return the old value (otherwise return null)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key or value is null, throw IllegalArgumentException
*/
@Override
public String put(String key, String value) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Remove the value corresponding to the given key from the data structure,
* if it exists. Return the old value, or null if no value was found.
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String remove(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* return the number of keys which start with the given prefix if the prefix
* contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public int countKeysMatchingPrefix(String prefix) {

// TODO Implement this, then remove this comment
return 0;
}
/*
* return the collection of keys which start with the given prefix if the
* prefix contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public List getKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Return the number of unique prefixes
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 8
* because the possible prefixes are G, GA, GAT, GATT, GATTC, GATTA, GATTAC,
GATTACA
*/
@Override
public int countPrefixes() {
// TODO Implement this, then remove this comment
return 0;
}
/*
* Return the sum of the lengths of all keys
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 15
*/
@Override
public int sumKeyLengths() {

// TODO Implement this, then remove this comment
return 0;
}
}
import java.util.List;
public interface PrefixMap {
public boolean isEmpty();
/*
* How many keys are stored in the map
*/
public int size();
/*
* Get the value corresponding to the key (or null if the key is not found)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
public String get(String key);
/*
* Insert the value into the data structure, using the given key. If the key
* already existed, replace and return the old value (otherwise return null)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key or value is null, throw IllegalArgumentException
*/
public String put(String key, String value);
/*
* Remove the value corresponding to the given key from the data structure,
* if it exists. Return the old value, or null if no value was found.
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
public String remove(String key);
/*
* return the number of keys which start with the given prefix if the prefix
* contains any character other than A,C,G,T, throw MalformedKeyException

* if the prefix is null, throw IllegalArgumentException
*/
public int countKeysMatchingPrefix(String prefix);
/*
* return the collection of keys which start with the given prefix if the
* prefix contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
public List getKeysMatchingPrefix(String prefix);
/*
* Return the number of unique prefixes
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 8
* because the possible prefixes are G, GA, GAT, GATT, GATTC, GATTA, GATTAC,
GATTACA
*/
public int countPrefixes();
/*
* Return the sum of the lengths of all keys
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 15
*/
public int sumKeyLengths();
}
Solution
import java.util.List;
import java.util.Map;
public class Storage implements PrefixMap {
private Map keys;
private Map> prefixes;
public Storage() {
// TODO Initialise your maps
}

/*
* How many keys are stored in the map
*/
@Override
public int size() {
// TODO Implement this, then remove this comment
return 0;
}
@Override
public boolean isEmpty() {
// TODO Implement this, then remove this comment
return false;
}
/*
* Get the value corresponding to the key (or null if the key is not found)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String get(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Insert the value into the data structure, using the given key. If the key
* already existed, replace and return the old value (otherwise return null)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key or value is null, throw IllegalArgumentException
*/
@Override
public String put(String key, String value) {
// TODO Implement this, then remove this comment

return null;
}
/*
* Remove the value corresponding to the given key from the data structure,
* if it exists. Return the old value, or null if no value was found.
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String remove(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* return the number of keys which start with the given prefix if the prefix
* contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public int countKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return 0;
}
/*
* return the collection of keys which start with the given prefix if the
* prefix contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public List getKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment

return null;
}
/*
* Return the number of unique prefixes
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 8
* because the possible prefixes are G, GA, GAT, GATT, GATTC, GATTA, GATTAC,
GATTACA
*/
@Override
public int countPrefixes() {
// TODO Implement this, then remove this comment
return 0;
}
/*
* Return the sum of the lengths of all keys
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 15
*/
@Override
public int sumKeyLengths() {
// TODO Implement this, then remove this comment
return 0;
}
}
import java.util.List;
public interface PrefixMap {
public boolean isEmpty();
/*
* How many keys are stored in the map
*/
public int size();

/*
* Get the value corresponding to the key (or null if the key is not found)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
public String get(String key);
/*
* Insert the value into the data structure, using the given key. If the key
* already existed, replace and return the old value (otherwise return null)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key or value is null, throw IllegalArgumentException
*/
public String put(String key, String value);
/*
* Remove the value corresponding to the given key from the data structure,
* if it exists. Return the old value, or null if no value was found.
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
public String remove(String key);
/*
* return the number of keys which start with the given prefix if the prefix
* contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
public int countKeysMatchingPrefix(String prefix);
/*
* return the collection of keys which start with the given prefix if the
* prefix contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
public List getKeysMatchingPrefix(String prefix);
/*
* Return the number of unique prefixes
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 8

* because the possible prefixes are G, GA, GAT, GATT, GATTC, GATTA, GATTAC,
GATTACA
*/
public int countPrefixes();
/*
* Return the sum of the lengths of all keys
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 15
*/
public int sumKeyLengths();
}

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Here is what I got so far, I dont know how to write union, interse.pdf

arihantpatna

META-INF/MANIFEST.MF Manifest-Version: 1.0 .classpath PriorityQueue.classpublicsynchronizedclass PriorityQueue { Heap q; public void PriorityQueue(int, java.util.Comparator); public Object peek(); public Object remove(); void add(Object); boolean isEmpty(); public int size(); } PriorityQueue.javaPriorityQueue.javaimport java.util.Comparator; publicclassPriorityQueue<E>{ Heap q; /** *PriorityQueue initializes the queue. * * @param initialCapacity an int that is the heaps initial size. * @param comparator the priority of various imputs. */ publicPriorityQueue(int initialCapacity,Comparator<?super E> comparator){ q=newHeap(initialCapacity,comparator); } /** * Peek, returns the next item in the queue without removing it. * * If it is empty then null is returned. * @return the next item in the queue. */ public E peek(){ if(q.size()==0){ returnnull; } return(E) q.findMax(); } /** * This removes the first item from the queue. * * It returns null if the queue is empty. * @return the first item in the queue. */ public E remove(){ if(q.size()==0){ returnnull; } return(E) q.removeMax(); } /** * This adds item to the queue * @param item that is added to the queue. */ void add(E item){ q.insert(item); } /** * isEmpty returns if the queue is empty or not. * * @return boolean if the queue is empty or not. */ boolean isEmpty(){ if(q.size()!=0){ returnfalse; } returntrue; } /** * size returns the size of the queue. * * @return int the size of the queue. */ publicint size(){ return q.size(); } } ArithmeticExpression.classpublicsynchronizedclass ArithmeticExpression { BinaryTree t; java.util.ArrayList list; String equation; void ArithmeticExpression(String) throws java.text.ParseException; public String toString(BinaryTree); public String toPostfixString(BinaryTree); void setVariable(String, int) throws java.rmi.NotBoundException; public int evaluate(BinaryTree); } ArithmeticExpression.javaArithmeticExpression.javaimport java.rmi.NotBoundException; import java.text.ParseException; import java.util.ArrayList; import java.util.Stack; /** * ArithmeticExpression takes equations in the form of strings creates a binary * tree, and can return either the regular or postfix equation. It also allows * them to be calculated. * * * Extra Credit: * ** it can handle spaces or no spaces in the string inputted. ** it can return * regular or postfix notation * * @author tai-lanhirabayashi * */ publicclassArithmeticExpression{ BinaryTree t; ArrayList list; String equation; /** * ArithmeticExpression is the construction which takes in a space * delimitated equation containing "*,/,+,-" symbols and converts it into a * binary tree. * * If the expression is not valid it will throw a ParseException. This is ...

META-INFMANIFEST.MFManifest-Version 1.0.classpath.docx

andreecapon

Google guava

Thomas Ferris Nicolaisen

Hi, Please find my code. I have correted all of your classes. Please let me know in case if do not get concpt. ######## Node.java ########### public class Node { private E element; private Node next; public Node() { this.element = null; this.next = null; } public Node(E e) { this.element = e; this.next = null; } public E getElement() { return this.element; } public void setElement(E element) { this.element= element; } public Node getNext() { return next; } public void setNext(Node next) { this.next = next; } } ########## Stack.java ############# public class Stack { private int N; private Node top; public Stack() { N = 0; this.top =null; } /* * places element on the top of the stack */ public void push(E element){ Node temp = new Node(element); temp.setNext(top); top = temp; N++; } /* * remove the top node and return its contents */ public E pop(){ if(top == null) return null; E e = top.getElement(); top = top.getNext(); N--; return e; } /* * Look at the top element of the Stack and return it, without removing */ public E peek(){ if(top == null) return null; return top.getElement(); } //returns the size of the stack public int size(){ return N; //replace } public boolean isEmpty(){ return top==null; } } ############ Queue.java ############ public class Queue { private Node front; private Node back; private int N; public Queue() { this.front = null; this.back = null; N = 0; } /* * places element in the back of the Queue */ public void enqueue(E element){ Node temp = new Node(element); if(front == null) { front = temp; back = temp; } else{ back.setNext(temp); back = temp; } N++; } /* * remove the front node of the queue and return it */ public E dequeue(){ if(front == null) { return null; } E item = front.getElement(); front = front.getNext(); // if we had only one element if(front == null){ back = null; } N--; return item; } /* * Look at the front of the queue and return it, without removing */ public E peek(){ if(back == null) return null; return back.getElement(); } //returns the size of the queue public int size(){ return N; } public boolean isEmpty(){ return front==null; } } ########### Palindrome.java ############ import java.util.Scanner; public class Palindrome { public static void main(String[ ] args) { Scanner input = new Scanner(System.in); String inputString; System.out.print(\"Please enter your string: \"); inputString = input.next( ); if (isPalindrome( inputString )){ System.out.println(\"Yes it is a palindrome.\"); } else{ System.out.println(\"No this is not a palindrome.\"); } } public static boolean isPalindrome(String input) { Queue q = new Queue (); Stack s = new Stack (); char letter; int i; for (i = 0; i < input.length( ); i++) { letter = input.charAt(i); q.enqueue(letter); s.push(letter); } while (!q.isEmpty( )) { if (q.dequeue() != s.pop( )) return false; } return true; } } /* Sample Output: Please enter your string: MadaM Yes it is a palindrome. */ Solution Hi, Please find my code. I have correted all of your classes. Pl.

Hi, Please find my code.I have correted all of your classes.Plea.pdf

pritikulkarni20

Please make it to clear 12-95. The basketball passed through the hoop even though it barely cleared the hands of the player B who attempted to block it. Neglecting the size of the ball, determine the magnitude vn of its initial velocity and the height h of the ball when it passes over player B 10 ft 5 ft Prob. 12-95 Solution Consider the motion along horizontal direction Vox = initial velocity in horizontal direction = Va Cos30 X = horizontal distance = 25 + 3 = 28 ft t = time taken t = X / Vox = 28/(Va Cos30) eq-1 consider the motion in vertical direction Y = vertical displacement = 10 - 7 = 3 ft a = acceleration = - 32.2 Voy = initial velocity = Va Sin30 t = time taken using the equation Y = Voy t + (0.5) a t2 3 = (Va Sin30) (28/(Va Cos30)) + (0.5) (- 32.2) (28/(Va Cos30))2 Va = 35.8 m/s t\' = time taken to reach B = 25 / (Va Cos30) = 25 / (35.8 Cos30) = 0.81 sec h = 7 + Voy t\' + (0.5) a t\'2 h = 7 + (35.8 Sin30) (0.81) + (0.5) (- 32.2) (0.81)2 h = 10.94 ft.

Please review my code (java)Someone helped me with it but i cannot.pdf

fathimafancyjeweller

Confitura 2012 Bad Tests, Good Tests

Tomek Kaczanowski

java question: \"Fill the add statement area\" Project is to work with stacks. package p2; public class Coordinate { public int x; public int y; public Coordinate( int x, int y ) { this.x = x; this.y = y; } public String toString() { return \"(\" + this.x + \",\" + this.y + \")\"; } @Override public boolean equals( Object object ) { if( object == null ) { return false; } if( ! Coordinate.class.isAssignableFrom( object.getClass() )) { return false; } final Coordinate other = (Coordinate) object; return this.x == other.x && this.y == other.y; } } package p2; public class Coordinate { public int x; public int y; public Coordinate( int x, int y ) { this.x = x; this.y = y; } public String toString() { return \"(\" + this.x + \",\" + this.y + \")\"; } @Override public boolean equals( Object object ) { if( object == null ) { return false; } if( ! Coordinate.class.isAssignableFrom( object.getClass() )) { return false; } final Coordinate other = (Coordinate) object; return this.x == other.x && this.y == other.y; } } package p2; import java.util.Vector; public class Maze { private char[][] maze; private int height; private int width; /** * Create a new Maze of the specified height and width, initializing every * location as empty, with a \' \'. **/ public Maze( int width, int height ) { // ADD STATEMENTS HERE } /** * Mutator to allow us to set the specified Coordinate as blocked, * marking it with a \'X\' **/ public void setBlocked( Coordinate coord ) { // ADD STATEMENTS HERE } /** * Mutator to allow us to set the specified Coordinate as having been visited, * marking it with a \'*\' **/ public void setVisited( Coordinate coord ) { // ADD STATEMENTS HERE } /** * Mutator to allow us to set the specified Coordinate as part of the path solution, * marking it with a \'.\' **/ public void setPath( Coordinate coord ) { // ADD STATEMENTS HERE } /** * Returns the character at the locatio specified by the Coordinate **/ public char at( Coordinate coord ) { // ADD STATEMENTS HERE } /** * Returns a Coordinate array containing all Coordinates that are clear around * the specified coordinate. **/ public Coordinate[] clearAround( Coordinate coord ) { Vector vector = new Vector(); // ADD STATEMENTS HERE // Look at each of the locations around the specified Coordinate, and add it // to the vector if it is clear (i.e. a space) return vector.toArray( new Coordinate[0] ); } /** * Returns a Coordinate that provides the entrance location in this maze. **/ public Coordinate start() { return new Coordinate( 0, 1 ); } /** * Returns a Coordinate that provides the exit location from this maze. **/ public Coordinate end() { // ADD STATEMENTS HERE } /** * The toString() method is responsible for creating a String representation * of the Maze. See the project specification for sample output. Note that * the String representation adds numbers across the top and side of the Maze * to show the Coordinates of each cell in the maze. **/ public String toString() { StringBuilder buffer =.

java question Fill the add statement areaProject is to wo.pdf

dbrienmhompsonkath75

How do I fix this error " Exception in thread "main" java.lang.NullPointerException: Cannot invoke "String.hashCode()" because "action" is null at AddressBook.main(AddressBook.java:25)" import java.util.Scanner; public class AddressBook { public static void main(String[] args) { Tablea addressBook = new Tablea(); Scanner scanner = new Scanner(System.in); String name, address; String action = null; boolean moreEntries = true; while (moreEntries) { System.out.println("/nMenu"); System.out.println("n.Add Entry"); System.out.println("d. Delete Entry"); System.out.println("u. Update Entry"); System.out.println("l. Search"); System.out.println("a. View All"); System.out.println("q. Exit"); // Insert a new entry into the address book switch (action) { case "n": System.out.print("Enter a name: "); name = scanner.nextLine(); System.out.print("Enter an address: "); address = scanner.nextLine(); addressBook.insert(name, address); System.out.println("New Contact has been added,"); System.out.print("Add another entry? (y/n): "); String answer = scanner.nextLine().toLowerCase(); moreEntries = answer.equals("y") || answer.equals("yes"); break; // Delete an entry from the address book case "d": System.out.println("Enter a name to delete: "); name = scanner.nextLine(); boolean deleted = addressBook.delete(name); if (deleted) { System.out.println("Address deleted"); } else { System.out.println("Name not found"); } break; // Update an entry in the address book case "u": System.out.print("Enter a name to update: "); name = scanner.nextLine(); System.out.print("Enter a new address: "); address = scanner.nextLine(); boolean updated = addressBook.update(name, address); if (updated) { System.out.println("Address updated"); } else { System.out.println("Name not found"); } break; // Lookup an entry in the address book case "l": System.out.print("Enter a name to look up: "); name = scanner.nextLine(); String result = addressBook.lookUp(name); if (result != null) { System.out.println("Address: " + result); } else { System.out.println("Name not found"); } break; // Display all entries in the address book case "a": System.out.println("All entries:"); addressBook.displayAll(); break; // Quit the program case"q": System.out.println("Quitting..."); } } } } Other files that go with it public class Node { private String name; private String address; private Node next; Node() { // add here .. } Node(String name, String address) { // add here .. this.name = name; this.address = address; this.next = null; } public String getKey() { // add here .. return this.name; } public void setKey(String name) { // add here .. this.name = name; } public String getValue() { // add here .. return this.address; } public void setValue(String address) { // add here .. this.address = address; } public Node getNext() { // add here .. return this.next; } public void setNext(Node next) { // add here .. this.next = next; } } import java.util.Scanner; public class Tablea { private Node mark; public Node.

How do I fix this error - Exception in thread -main- java-lang-NullPoi.pdf

pnaran46

// main public class AssignmentThree { public static void main(String[] args) { List myList = new List(15); // Cause List Empty Message myList.removeFront(); myList.removeRear(); myList.removeItem(\"a\"); // Cause Not found message myList.addToFront(\"x\"); myList.removeItem(\"y\"); myList.removeItem(\"x\"); myList.addAfterItem(\"x\", \"z\"); myList.addBeforeItem(\"x\", \"z\"); // Normal behavior myList.addToFront(\"not.\"); myList.addToFront(\"or\"); myList.addToRear(\"is\"); myList.addToRear(\"try.\"); myList.addAfterItem(\"is\", \"no\"); myList.addBeforeItem(\"is\", \"There\"); myList.addToFront(\"Do\"); myList.addAfterItem(\"or\", \"do\"); myList.print(\"Original list\"); myList.printSorted(\"Sorted Original List\"); sop(\"\ Front is \" + myList.getFront()); sop(\"Rear is \" + myList.getRear()); sop(\"Count is \" + myList.askCount()); sop(\"Is There present? \" + myList.isPresent(\"There\")); sop(\"Is Dog present? \" + myList.isPresent(\"Dog\")); myList.addToFront(\"junk\"); myList.addToRear(\"morejunk\"); myList.addAfterItem(\"or\", \"moremorejunk\"); myList.print(\"List with junk\"); sop(\"Count is \" + myList.askCount()); myList.removeFront(); myList.removeRear(); myList.removeItem(\"moremorejunk\"); myList.print(\"List with junk removed\"); sop(\"Count is \" + myList.askCount()); sop(\"\"); // Cause List Full message for(int ii = 0; ii < 10; ++ii) { myList.addToFront(DUMMY); } myList.addToRear(DUMMY); myList.addBeforeItem(\"no\", DUMMY); myList.addAfterItem(\"There\", DUMMY); myList.print(\"After filling List\"); sop(\"Count is \" + myList.askCount()); while(myList.isPresent(DUMMY)) myList.removeItem(DUMMY); myList.print(\"After removing \" + DUMMY ); sop(\"Count is \" + myList.askCount()); } private static void sop(String s) { System.out.println(s); } private static final String DUMMY = \"dummy\"; } Solution Following is MyList implementation for your output as: Java Code: import java.util.Comparator; public class MyList { private Node start; private Node end; private int size; private int capacity; public MyList(int capacity) { start = null; end = null; size = 0; this.capacity = capacity; } /* Function to check if list is empty */ public boolean isEmpty() { return start == null; } public void removeFront() { if (isEmpty()) { System.out.println(\"List Empty\"); return; } else { start = start.next; size--; return; } } public void removeRear() { if (size == 0) { System.out.println(\"List Empty\"); return; } if (start.next == null) { start=null; end=null; size=0; } Node node = start.next; while (node.next != null) { node = node.next; } node.next = null; end = node; size--; } public void removeItem(String key) { if (size == 0) { System.out.println(\"List Empty\"); return; } if (start.next == null) { if (start.key.equals(key)) { start=null; end=null; size=0; } else { System.out.println(\"\"); } return; } Node node = start; Node nodeNext = node.next; while (nodeNext.next != null) { if (nodeNext.key.equals(key)) { node.next=nodeNext.next; }.

mainpublic class AssignmentThree { public static void ma.pdf

fathimafancyjeweller

GeeCON 2012 Bad Tests, Good Tests

Tomek Kaczanowski

in this assignment you are asked to write a simple driver program and set of functions (maybein a library) that can be performed on a binary search tree. Your program should allow user to insert/delete integer values into the binary search tree along with several other operations on the binary search tree. You can use the code given in slides. But this time your key will be int! Specifically, your program will ask user to enter a command and related parameters (if any) in a loop, and then perform the given commands. Here is the list of commands that your program must implement: * insert *find\' *delete *list inorder *list preorder *list postorder *list levelorder * max * min * height *count * sum *quit As always, make sure you release (free) the dynamically allocated memories if you allocate any memory in your programs. So, before submitting your program, run it with valgrind to see if there is any memory leakage //my proggram in C struct tree_node { int data; struct tree_node *left, *right; } typedef struct nodeT { int key; struct nodeT *left, *right; } nodeT, *treeT; int main(){ while (TRUE) { printf(\"> \"); line = GetLine(); ch = toupper(line[0]); switch (ch) { case \'I\': insert(); break; case \'F\': find(); break; case \'D\': delete(); break; case \'LI\': listInorder; break; case \'LPR\': listPreorder(); break; case \'LPO\': listPostorder(); break; case \'MAX\': max(); break; case \'min\': min(); break; case \'H\': height(); break; case \'C\': count(); break; case \'S\': sum(); break; case \'Q\': exit(0); default:printf(\"Illegal command\ \"); break; } } } nodeT *FindNode(nodeT *t, int key){ while(t !=NULL) { if (key == t->key) return t; if (key < t->key) { t = t->left; } else { t = t->right; } return NULL; } void delete(nodeT **p){ nodeT *target; target=*p; if (target->left==NULL && target->right==NULL) { *p=NULL; } else if (target->left == NULL) { *p=target->right; } else if (target->right == NULL) { *p=target->left; } else { /* target has two children, see next slide */ } free(target); } void listInorder(nodeT *T){ if (t != NULL) { DisplayTree(t->left); printf(“%d “, t->key); DisplayTree(t->right); } } void listPreorder(nodeT *t) { if (t != NULL) { printf(“%d “, t->key); DisplayTree(t->left); DisplayTree(t->right); } } void listPostOrder(nodeT *t){ if (t != NULL) { DisplayTree(t->left); DisplayTree(t->right); printf(“%d “, t->key); } } void intsert(nodeT **tptr, int key){ nodeT*t, *tmp; t=*tptr; if (t == NULL) { tmp=New(nodeT*); tmp->key = key; tmp->left=tmp->right=NULL; *tptr=tmp; return; } if (key < t->key) { InsertNode (&t->left, key); } else { InsertNode(&t->right, key); } } int height(nodeT *t){ if (t == NULL) return 0; else return (1 + maximumof( height(t->left), height(t->right)) ); } int sum(struct tree_node *p){ if (p == NULL) return 0; else return (p->data + sum(p->left) + sum(p->right) ); } Solution 1. /* 2. * Java Program to Implement Binary Search Tree 3. */ 4. 5. import java.util.Scanner; 6. 7. /* Class BSTNode */ 8. cl.

in this assignment you are asked to write a simple driver program an.pdf

michardsonkhaicarr37

How to fix this error: Exception in thread "main" q. Exit java.lang.NullPointerException: Cannot invoke "String.hashCode()" because "action" is null at AddressBook.main(AddressBook.java:23) import java.util.Scanner; public class AddressBook { public static void main(String[] args) { Tablea addressBook = new Tablea(); Scanner scanner = new Scanner(System.in); String name, address; String action = null; boolean moreEntries = true; while (moreEntries) { System.out.println("/nMenu"); System.out.println("n.Add Entry"); System.out.println("d. Delete Entry"); System.out.println("u. Update Entry"); System.out.println("l. Search"); System.out.println("a. View All"); System.out.println("q. Exit"); // Insert a new entry into the address book switch (action) { case "n": System.out.print("Enter a name: "); name = scanner.nextLine(); System.out.print("Enter an address: "); address = scanner.nextLine(); addressBook.insert(name, address); System.out.println("New Contact has been added,"); System.out.print("Add another entry? (y/n): "); String answer = scanner.nextLine().toLowerCase(); moreEntries = answer.equals("y") || answer.equals("yes"); break; // Delete an entry from the address book case "d": System.out.println("Enter a name to delete: "); name = scanner.nextLine(); boolean deleted = addressBook.delete(name); if (deleted) { System.out.println("Address deleted"); } else { System.out.println("Name not found"); } break; // Update an entry in the address book case "u": System.out.print("Enter a name to update: "); name = scanner.nextLine(); System.out.print("Enter a new address: "); address = scanner.nextLine(); boolean updated = addressBook.update(name, address); if (updated) { System.out.println("Address updated"); } else { System.out.println("Name not found"); } break; // Lookup an entry in the address book case "l": System.out.print("Enter a name to look up: "); name = scanner.nextLine(); String result = addressBook.lookUp(name); if (result != null) { System.out.println("Address: " + result); } else { System.out.println("Name not found"); } break; // Display all entries in the address book case "a": System.out.println("All entries:"); addressBook.displayAll(); break; // Quit the program case"q": System.out.println("Quitting..."); } } } } OTHER FILES THAT GO WITH THIS public class Node { private String name; private String address; private Node next; Node() { // add here .. } Node(String name, String address) { // add here .. this.name = name; this.address = address; this.next = null; } public String getKey() { // add here .. return this.name; } public void setKey(String name) { // add here .. this.name = name; } public String getValue() { // add here .. return this.address; } public void setValue(String address) { // add here .. this.address = address; } public Node getNext() { // add here .. return this.next; } public void setNext(Node next) { // add here .. this.next = next; } } import java.util.Scanner; public class Tablea { private Node mark; publi.

How to fix this error- Exception in thread -main- q- Exit java-lang-.pdf

aarokyaaqua

Implement the ADT stack by using an array stack to contain its entries. Expand the array dynamically, as necessary. Maintain the stack’s bottom entry in stack[stack.length – 1]. Your Stack class must implement StackInterface (provided). You may use ArrayStack.java (provided) as the starting point for your implementation. You must implement a comprehensive set of unit tests using the main() method (and private utility methods) in ArrayStack.java. ArrayStack.java public class ArrayStack implements StackInterface { private T[] stack; // Array of stack entries private int topIndex; // Index of top entry private boolean initialized = false; private static final int DEFAULT_CAPACITY = 50; private static final int MAX_CAPACITY = 10000; public ArrayStack() { this(DEFAULT_CAPACITY); } // end default constructor public ArrayStack(int initialCapacity) { checkCapacity(initialCapacity); // The cast is safe because the new array contains null entries @SuppressWarnings(\"unchecked\") T[] tempStack = (T[])new Object[initialCapacity]; stack = tempStack; topIndex = -1; initialized = true; } // end constructor // < Implementations of the stack operations go here. > // < Implementations of the private methods go here; checkCapacity and // checkInitialization are analogous to those in Chapter 2. > // . . . } // end ArrayStack StackInterface.java public interface StackInterface { /** Adds a new entry to the top of this stack. @param newEntry An object to be added to the stack. */ public void push(T newEntry); /** Removes and returns this stack\'s top entry. @return The object at the top of the stack. @throws EmptyStackException if the stack is empty before the operation. */ public T pop(); /** Retrieves this stack\'s top entry. @return The object at the top of the stack. @throws EmptyStackException if the stack is empty. */ public T peek(); /** Detects whether this stack is empty. @return True if the stack is empty. */ public boolean isEmpty(); /** Removes all entries from this stack. */ public void clear(); } // end StackInterface Solution StackTester.java public class StackTester { public static void main(String[] args) { ArrayStack aS = new ArrayStack(); aS.push(3); aS.push(5); aS.push(10); aS.push(11); System.out.println(\"\"+ aS.peek() + \", \" + aS.pop() + \", \" + aS.peek2()); aS.remove(1); } } ArrayStack.java import java.util.Arrays; public class ArrayStack implements StackInterface { private T[] stack; private int topIndex; private boolean initialized = false; private static final int DEFAULT_CAPACITY = 50; private static final int MAX_CAPACITY = 10000; public ArrayStack() { this(DEFAULT_CAPACITY); } public ArrayStack(int initialCapacity) { checkCapacity(initialCapacity); @SuppressWarnings(\"unchecked\") T[] tempStack = (T[])new Object[initialCapacity]; stack = tempStack; topIndex = -1; initialized = true; } public boolean isEmpty() { return topIndex < 0; } private void ensureCapacity() { if(topIndex == stack.length - 1) { int newLength = 2*stack.length; .

Implement the ADT stack by using an array stack to contain its entri.pdf

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mainpublic class AssignmentThree { public static void ma.pdf

GeeCON 2012 Bad Tests, Good Tests

in this assignment you are asked to write a simple driver program an.pdf

How to fix this error- Exception in thread -main- q- Exit java-lang-.pdf

Implement the ADT stack by using an array stack to contain its entri.pdf

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What function can be used to return multiple numeric data such as in.pdf

sauravmanwanicp

Answer the following questions in reference to Drosophila melanogaster (fruit fly) that has 6 chromosomes. How many combinations are possible when chromosomes assort independently into gametes (ignoring crossing over)? 6 raised to the 2nd power 2 raised to the 6th power 6 x 2 = 12 2 raised to the 3rd power 3 raised to the 2nd power How many tetrads align at the metaphase plate of meiosis I? 0 12 6 3 Solution Ans.1 No. of possible combination during independent assortment = 2n Where, 2 indicates ploidy of the somatic cells undergoing meiosis, n = number of haploid chromosomes. Given, 2n = 6 Hence, n = 3. Thus, No. of possible combination during independent assortment = 2n = 23 Correct option: 2 raised to 3rd power. Ans. 2. Number of tetrad during meiosis = Number of haploid chromosomes (n) To understand it, let’s number the three haploid chromosomes as A, B and C. A diploid cell has two copies of each chromosome- one inherited from mother and the other from father. So, diploid cell has following chromosomes- A1, A2, B1, B2, C1 and C3, where 1= maternal chromosomes and 2 = fraternal chromosomes. All the chromosomes remain independent of each other, randomly scattered in nucleus. We will proceed with one chromosome pair, for simplicity. The following events occur when a diploid cell enters meiosis- A1 and A2 are called homologues pair. During S-phase, DNA is duplicated. The resultant- I. A1A1 remain joined to each other. Are called sister chromatids. The same happens to A2A2, and rest other chromosomes. II. Homologous chromosome A1 and A2 pair together prophase I. Since each of A1 and A2 are duplicated, they are (A1A1) and (A2A2) respectively. III. Homologous pairing forms a tetrad – A1A1/A2A2 i.e. four chromosomes remaining together. Note: Both the maternal and fraternal chromosomes and their duplicated copies remain together as a tetrad. That is, the number of tetrad in half of diploid number (because A1 from mother and A2 from father remain together as one single homologous chromosome). Thus, number of treads in drosophila = n = 3 Correct option: D (3).

Answer the following questions in reference to Drosophila melanogast.pdf

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There are five feasible solutions to a three-objective optimization p.pdf

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An air-filled parallel-plate capacitor has capacitance Co. If two identically sized dielectric slabs of dielectric constants K_1 and K_2 are inserted as shown in the following figure, what is the new capacitance? (Hnt: treat this as two capacitor in combinations) Solution it would act as capacitors in parallel. C= K0A/d. Non C1= A/(d/2)*K1=2C/k1 Simlarly C2= 2C/K2 Since it can be treated as capacitor is parallel so Cfinal= 2C/K1 + 2c/k2 = 2C(K1+K2/K1*K2).

An air-filled parallel-plate capacitor has capacitance Co. If two ide.pdf

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The incandescent lightbulb was actually invented by Humphry Davy in 1802. It was not commercializable in that form. Joseph Swan was successful in making a commercializable unit in the late 1870s. He may have been granted the first light bulb patent (British patent 4933, issued in 1880). Thomas Edison was also developing electric lighting at that time. He applied for US patents as early as 1878 So why is Edison credited as the inventor? Partly because he was the first to design and manufacture a robust lighting system having a bulb with an effective incandescent material, higher vacuum inside the bulb, and high resistance so power distribution from a central source was possible. Perhaps more importantly was the development ofsupply sources and materials for distribution networks. A modem incandescent lightbulb is shown below. Overall efficiency, defined as light power out divided by electrical power in, is about 2.5% for a 100 W unit. 1 1. outline of glass bulb 2. Low pressure inert gas (argon, nitrogen, krypton, xenon) 3. Tun ten filament 5) 4. Contact wire (goes out of stem) 5. Contact wire (goes into stem) 6. Support wires (one end embedded in stem; conduct no current) 7. Stem (glass mount) 8. Contact wire (goes out of Identify five different systems that could be used for an energy analysis (some combination Draw arrows for each systems\' energy flow process(es)-W Qooeduction, Qconvection, radiation. Solution To conduction and convection of heat must have the medium. in vaccum only radiation is there. given problem the heat coming out from the tungsten filament(3). the work supplied by the electrical current through the wires. first humphry davy invented incandescent bulb having vaccum inside so there is no possibilty of heat transfer through conduction or convection. next edison filled with the low pressure inert gases. beacuse of that from hot Tungsten filamet to inert gases heat transfer occurs through convection. next inertgasses to glass or stem heated by the convection (fluid and solid interaction) again glass to sourroundings is convection(solid and fluid interaction) radiation is always there beacuse it possible with medium and with out medium contact wires are heated up through the conduction..

The incandescent lightbulb was actually invented by Humphry Davy in 1.pdf

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The fossil record provides little information about ancient mosses. Do you think nonvascular plants could have ever been as large as trees or shrubs? Provide evidence based on your knowledge of nonvascular plant anatomy. Solution 2). No, the nonvascular plants could have never been as large as trees or shrubs because they cannot provide the water and nutrients to the plant parts that are away from the source of water. The nonvascular plants cannot control their water supply internally, it is solely depends upon their external environment that provides water to the plant. So, they most likely grown over a lake or rocks to get the water and propagated vegetative (clonal moss)..

The fossil record provides little information about ancient mosses. D.pdf

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The DNA sequence below is a shortened version of the sequence on the previous page. This DNA sequence corresponds to the mRNA sequence only; two introns are not included. The numbers indicate the position of nucleotides within the sequence and indicates intervening nucleotides that are not shown. The gene is 626 nucleotides long. Recall that chromosomes are very long DNA molecules with millions of nucleotides, so this sequence is preceded and followed by seemingly endless sequences of nucleotides. IDNA is double-stranded. Write the sequence of the first 10 nucleotides of the complementary strand of DNA. Solution ans. DNA is a double stranded which consist of two strands and one strand is complemetary to the other.DNA contains complemetary base pairs in which adenine is linke d to tymine by two hydrogen bonds guanine to cytosine by three hydrogen bonds The seque of complementary strand of the DNA molecule is 5\' TGTAAACGAA.

The DNA sequence below is a shortened version of the sequence on the .pdf

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the conducting tissue found in vascular plants that functions in transportation of materials throughout the plant, but does not lend structure support to the stem is A. xylem B. phloem C.sporopollenin D.endosperm e. LYCOPHYLL Solution Answer : B. Phloem Phloem is responsible for the transporting sugars,proteins, and other organic molecules in plants. xylem is responsible for the transport of water and water soluble nutrients,taken up by the roots and also primary component of wood. Vascular plants are able to higher than other plants due to the rigidity of xylem cells, which support the plant..

the conducting tissue found in vascular plants that functions in tra.pdf

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Show that f(x)=57^x+1 and f^-1(x)=7x-75 are inverses Choose the ap.pdf

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Read the short article Horizontal Gene Transfer in E Coli and answer the following questions.(https://www.sciencedaily.com/releases/2015/05/150519105900.htm) Remember to use your own words; plagiarism of any kind will not be tolerated (see syllabus for further information on what constitutes plagiarism). (1) What is horizontal gene transfer (HGT)? (2) In the posted article, how many gene transfer elements were found for E. coli O104:H4? (3) What is the source of these gene transfer elements? (4) Should we be concerned about HGT in prokaryotes? If so, why? Solution (1) Horizontal gene transfer (HGT) is the lateral gene movement, wherein the movement of genetic material takes place between unicellular or multicellular organisms across species. This gene transfer is different from the vertical gene transfer in which the transmission of DNA occurs from parent to offspring. Often genetic material swaping occurs between bacteria of same as well as different species, through this means by aquiring one of the processes; transformation, transduction or conjugation. During HGT, plasmids, transposons or integrons act as vectors that transfer genes. (2) In the given article, the team from Madurai Kamaraj University in Madurai, Tamil Nadu, identified 38 gene transfer elements for E. coli O104:H4. (3) The research team identified 38 gene transfer prophage elements for E.coli O104:H4. They showed that the these elements act as genetic weapons protecting the bacteria from antibiotics. The source of these elements were viruses (the bacteriophages), that infect bacteria, and in the process bacteria aquired these elements from viruses. (4) In prokaryotes, horizontal gene transfer is a major concern. Aquisition of new genes through HGT is ability of Bacteria and Archaea to adapt to new changing environments. Druing HGT, genes that are responsible for antibiotic resistance in one species of bacteria can be transferred to another species of bacteria, and is the primary reason for the spread of antibiotic resistance. Besides antibiotic resistance, HGT also gives bacteria an ability to degrade novel compounds (such as different insecticides and pesticides). Thus the bacterial strains after HGT are much more dangerous as they are multi drug resistant and more virulent. These transfer elements when transferred to a pathogenic bacteia, can become a serious health hazard as they are resistant to antibiotics available with us and thus cannot be trated easily..

Read the short article Horizontal Gene Transfer in E Coli and answer.pdf

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Please select the best answer and click submit. The owners of fou.pdf

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One function of the spliceosome is to: (Select all that apply.) A. recognize the beginning and end of the introns. B. recognize the 5\' end of the primary transcript. C recognize the beginning and end of the exons. D recognize the 3\' end of the primary transcript. Solution The correct answer is :A. recognize the beginning and end of the introns. Reason: Spliceosome consists of snRNAs and snRNPs which together make a complex.This complex binds to the beginning and end of the introns and forms a loop which is later excised and the ends of exons are sealed..

One function of the spliceosome is to (Select all that apply.)A. .pdf

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Match the following organelles with their key functions in the cell nucleus rough endoplasmic reticulum Golgi apparatus lysosomes mitochondria chloroplasts peroxisomes cytosol carbon fixation site of protein synthesis ready for cellular distribution recycling of intracellular materials modification and packing of proteins for secreation site of protein synthesis that contains the majority of metabolic pathways containment of toxic reactions the site of the core genome oxidative phosphorylation Solution 1: Nucleus: The site of core genome;g 2: Rough ER: site of protein synthesis ready for cellular distribution; b 3: Golgi apparatus: Modification and packing of proteins for secretion; d 4: Lysosome: recylcing of intracellular material; c 5: Mitochondria: Oxidative phosphorylation;h 6: Chloroplast: Carbon fixation; a 7: Peroxisomes: Containment of toxic reaction; f 8: Cytosol: Site of protein synthesis that contains majority of metabolic pathways; e.

Match the following organelles with their key functions in the cell .pdf

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please choose the correct answerIn the ABO blood-type phenotype, w.pdf

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mo Press Submit until all questions are colored green and you have do.pdf

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matching: chose the stage of the light reactions which the statement occurs in. 1. _Produces ATP. 2. _Passes energized electrons down an electron transport chain to create ATP. 3. _Produces NADPH. 4. _Splits water to get electrons to replace energized electrons. 5. _Replaces energized electrons from electrons that have completed the 1st electron transport chain. a. water-splitting photosystem. b. NADPH-Producing photosystem. Solution During water splitting photosystem ,the electrons in water are removed and passed to the chlorophyll. The hydrogen ions extracted from the water molecule, and after two water molecules are splitted,the oxygen atoms takes electron and form molecular oxygen (O 2 ). In photosynthesis, water splitting donates electrons to the electron transport chain in photosystem II. So it splits water to get electrons to replace energized electrons. In the photosystem II (PSII) reaction center, energy from sunlight is used to extract electrons from water. The electrons travel via the chloroplast electron transport chain to photosystem I (PSI), which reduces NADP+ to NADPH. So it produces NADPH.

matching chose the stage of the light reactions which the statement.pdf

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Let E be the set of all even integers, and let O be the set of all odd integers. Explain why Z E O. A. Every integer is both even and odd. B. Some integers are both even and odd. C. Even integers and odd integers are integers. D. Every integer is either even or odd. E. No integer is both even and odd. Solution ans : D take any integer z, z is either odd or even so, for all z , Z is a subset of EUO.

Let E be the set of all even integers, and let O be the set of all o.pdf

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Indexes prevent the database engine from having to scan every record in the database. True False Solution Answer: True Indexes prevent the database engine from having to scan every record in the database. Indexes provides the improvement of query performance by providing the database engine with fast way of locating a particular record based on the a column or columns provided in where condition..

Indexes prevent the database engine from having to scan every record .pdf

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