Need this in JAVA We have N numbers as an array, you need to find a .pdf

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Need this in JAVA We have N numbers as an array, you need to find a prefix array and a suffix array, which we can get the maximum x or value with all elements in them. Notice that for prefix[0, 1] and suffix[r, n - 1], do not intersect (I Solution // C++ program for a Trie based O(n) solution to find max // subarray XOR #include using namespace std; // Assumed int size #define INT_SIZE 32 // A Trie Node struct TrieNode { int value; // Only used in leaf nodes TrieNode *arr[2]; }; // Utility function tp create a Trie node TrieNode *newNode() { TrieNode *temp = new TrieNode; temp->value = 0; temp->arr[0] = temp->arr[1] = NULL; return temp; } // Inserts pre_xor to trie with given root void insert(TrieNode *root, int pre_xor) { TrieNode *temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[val] == NULL) temp->arr[val] = newNode(); temp = temp->arr[val]; } // Store value at leaf node temp->value = pre_xor; } // Finds the maximum XOR ending with last number in // prefix XOR \'pre_xor\' and returns the XOR of this maximum // with pre_xor which is maximum XOR ending with last element // of pre_xor. int query(TrieNode *root, int pre_xor) { TrieNode *temp = root; for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[1-val]!=NULL) temp = temp->arr[1-val]; // If there is no prefix with opposite // bit, then look for same bit. else if (temp->arr[val] != NULL) temp = temp->arr[val]; } return pre_xor^(temp->value); } // Returns maximum XOR value of a subarray in arr[0..n-1] int maxSubarrayXOR(int arr[], int n) { // Create a Trie and insert 0 into it TrieNode *root = newNode(); insert(root, 0); // Initialize answer and xor of current prefix int result = INT_MIN, pre_xor =0; // Traverse all input array element for (int i=0; i using namespace std; // Assumed int size #define INT_SIZE 32 // A Trie Node struct TrieNode { int value; // Only used in leaf nodes TrieNode *arr[2]; }; // Utility function tp create a Trie node TrieNode *newNode() { TrieNode *temp = new TrieNode; temp->value = 0; temp->arr[0] = temp->arr[1] = NULL; return temp; } // Inserts pre_xor to trie with given root void insert(TrieNode *root, int pre_xor) { TrieNode *temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[val] == NULL) temp->arr[val] = newNode(); temp = temp->arr[val]; } // Store value at leaf node temp->value = pre_xor; } // Finds the maximum XOR ending with last number in // prefix XOR \'pre_xor\' and returns the XOR of this maximum // with pre_xor which is maximum XOR ending with last element // of pre_xor. int query(TrieNode *root, int pre_xor) { TrieNode *temp = root; for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[1-val]!=NULL) temp = temp-.

Education

$Need this in JAVA We have N numbers as an array, you need to find a prefix array and a suffix array, which we can get the maximum x or value with all elements in them. Notice that for prefix[0, 1] and suffix[r, n - 1], do not intersect (I Solution // C++ program for a Trie based O(n) solution to find max // subarray XOR #include using namespace std; // Assumed int size #define INT_SIZE 32 // A Trie Node struct TrieNode { int value; // Only used in leaf nodes TrieNode *arr[2]; }; // Utility function tp create a Trie node TrieNode *newNode() { TrieNode *temp = new TrieNode; temp->value = 0; temp->arr[0] = temp->arr[1] = NULL; return temp; } // Inserts pre_xor to trie with given root void insert(TrieNode *root, int pre_xor) { TrieNode *temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix$

$bool val = pre_xor & (1<arr[val] == NULL) temp->arr[val] = newNode(); temp = temp->arr[val]; } // Store value at leaf node temp->value = pre_xor; } // Finds the maximum XOR ending with last number in // prefix XOR 'pre_xor' and returns the XOR of this maximum // with pre_xor which is maximum XOR ending with last element // of pre_xor. int query(TrieNode *root, int pre_xor) { TrieNode *temp = root; for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[1-val]!=NULL) temp = temp->arr[1-val]; // If there is no prefix with opposite // bit, then look for same bit. else if (temp->arr[val] != NULL) temp = temp->arr[val]; } return pre_xor^(temp->value); } // Returns maximum XOR value of a subarray in arr[0..n-1] int maxSubarrayXOR(int arr[], int n) { // Create a Trie and insert 0 into it TrieNode *root = newNode(); insert(root, 0); // Initialize answer and xor of current prefix int result = INT_MIN, pre_xor =0; // Traverse all input array element for (int i=0; i$

$using namespace std; // Assumed int size #define INT_SIZE 32 // A Trie Node struct TrieNode { int value; // Only used in leaf nodes TrieNode *arr[2]; }; // Utility function tp create a Trie node TrieNode *newNode() { TrieNode *temp = new TrieNode; temp->value = 0; temp->arr[0] = temp->arr[1] = NULL; return temp; } // Inserts pre_xor to trie with given root void insert(TrieNode *root, int pre_xor) { TrieNode *temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[val] == NULL) temp->arr[val] = newNode(); temp = temp->arr[val]; } // Store value at leaf node temp->value = pre_xor; } // Finds the maximum XOR ending with last number in // prefix XOR 'pre_xor' and returns the XOR of this maximum // with pre_xor which is maximum XOR ending with last element$

$// of pre_xor. int query(TrieNode *root, int pre_xor) { TrieNode *temp = root; for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[1-val]!=NULL) temp = temp->arr[1-val]; // If there is no prefix with opposite // bit, then look for same bit. else if (temp->arr[val] != NULL) temp = temp->arr[val]; } return pre_xor^(temp->value); } // Returns maximum XOR value of a subarray in arr[0..n-1] int maxSubarrayXOR(int arr[], int n) { // Create a Trie and insert 0 into it TrieNode *root = newNode(); insert(root, 0); // Initialize answer and xor of current prefix int result = INT_MIN, pre_xor =0; // Traverse all input array element for (int i=0; i$

Similar to Need this in JAVA We have N numbers as an array, you need to find a .pdf

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Write a program that accepts an arithmetic expression of unsigned integers in postfix notation and builds the arithmetic expression tree that represents that expression. From that tree, the corresponding fully parenthesized infix expression should be displayed and a file should be generated that contains the three address format instructions. This topic is discussed in the week 4 reading in module 2, section II-B. The main class should create the GUI shown below: Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the corresponding infix expression should be displayed and the three address instruction file should be generated. The postfix expression input should not be required to have spaces between every token. Note in the above example that 9+- are not separated by spaces. The above example should produce the following output file containing the three address instructions: Add R0 5 9 Sub R1 3 R0 Mul R2 2 3 Div R3 R1 R2 Inheritance should be used to define the arithmetic expression tree. At a minimum, it should involve three classes: an abstract class for the tree nodes and two derived classes, one for operand nodes and another for operator nodes. Other classes should be included as needed to accomplish good object-oriented design. All instance data must be declared as private. You may assume that the expression is syntactically correct with regard to the order of operators and operands, but you should check for invalid tokens, such as characters that are not valid operators or operands such as 2a, which are not valid integers. If an invalid token is detected a RuntimeException should be thrown and caught by the main class and an appropriate error message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3 Construct Tree Infix Expression ((3-(5 9))/ (2* 3 Solution #include #include #include #include #include struct TREE //Structure to represent a tree { char data; struct TREE *right,*left,*root; }; typedef TREE tree; /* Stack Using Linked List */ struct Stack //Structure to represent Stack { struct TREE *data; struct Stack *next,*head,*top; //Pointers to next node,head and top }; typedef struct Stack node; node *Nw; //Global Variable to represent node void initialize(node *&n) { n = new node; n -> next = n -> head = n -> top = NULL; } void create(node *n) { Nw = new node; //Create a new node Nw -> next = NULL; //Initialize next pointer field if(n -> head == NULL) //if first node { n -> head = Nw; //Initialize head n -> top = Nw; //Update top } } void push(node *n,tree *ITEM) { node *temp = n -> head; if(n -> head == NULL) //if First Item is Pushed to Stack { create(n); //Create a Node n -> head -> data = ITEM; //Insert Item to head of List n -> head = Nw; return; //Exit from Function } create(n); //Create a new Node Nw -> data = ITEM; //Insert Item while(temp -> next != NULL) temp = temp -> next; //Go to Last Node temp -> next = Nw; //Point New node n -> top = Nw; //Upda.

Write a program that accepts an arithmetic expression of unsigned in.pdf

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I know that one of the main differences between the rough and smooth endoplasmic reticulum is the presence or absence of ribosomes. The structures also have different shapes, rough ER is flat, and the smooth ER is more round. My question is what specific process causes the two different forms of ER to divide and be so different?At what point does the ER decide to stop making ribosomes and begin to initiate the synthesis of smooth ER? Solution Answer :- The rough ER is specialized with function of protein synthesis and smooth ER with lipid synthesis. When cell is need of lipids and need less proteins the rough ER undergoes loss of ribosomes and gets transformed into smooth ER. ER are actually the membrane extensions of nuclear memebrane..

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Modify the following source code so that when the mouse is clicked within 30 pixels of the middle of the smiley face it changes color. It will alternate between the yellow and red smiley faces. two images have been given to you , redSmiley.gif and happyFace.gif to get started. Just for fun: Create more than 2 colors of smiley faces and rotate through the images changing each time the mouse is clicked within the threshold. Use different images other than smiley faces, change images each time the mouse is clicked within the threshold. Add a sound each time the image changes. Have the image slow down, if it has been more than x number of seconds since the image has changed. Solution import java.awt.*; import java.awt.event.*; import java.awt.image.BufferedImage; import javax.swing.*; public class ChangeImageOnClick { private void showContentsImage() { JFrame driver = new JFrame(\"Move pointer on Image\"); driver.setDefaultCloseOperation(JFrame.DISPOSE_ON_CLOSE); CustomPanel contentPane = new CustomPanel(); driver.setContentPane(contentPane); driver.pack(); driver.setLocationByPlatform(true); driver.setVisible(true); } public static void main(String\\u005B\\u005D args) { SwingUtilities.invokeLater(new Runnable() { public void run() { new ChangeImageOnClick().showContentsImage(); } }); } } class CustomPanel extends JComponent { private final int SIZE = 50; private int imageX = 100; private int imageY = 100; private int imageIndex; private ImageIcon image; private ImageIcon firstImage; private ImageIcon secondImage; private java.net.URL url; private Rectangle boundsForMouse; public CustomPanel() { image = new ImageIcon(); try { url = new java.net.URL(\"please mention url location of imaage\"); firstImage = new ImageIcon(url); url = new java.net.URL(\"please mention url location of image\"); secondImage = new ImageIcon(url); } catch(Exception e) { e.printStackTrace(); } imageIndex = 1; image.setImage(firstImage.getImage()); boundsForMouse = new Rectangle(imageX - 30, imageY - 30, firstImage.getIconWidth() + 60, firstImage.getIconHeight() + 60); setOpaque(true); addMouseListener(new MousePosition()); } private int setImage(int counter) { System.out.println(\"Image Index : \" + counter); if (counter == 1) { image = new ImageIcon(); image.setImage(secondImage.getImage()); boundsForMouse = new Rectangle(imageX - 30, imageY - 30, secondImage.getIconWidth() + 60, secondImage.getIconHeight() + 60); repaint(); counter++; return (counter); } else if (counter == 2) { image = new ImageIcon(); image.setImage(firstImage.getImage()); boundsForMouse = new Rectangle(imageX - 30, imageY - 30, firstImage.getIconWidth() + 60, firstImage.getIconHeight() + 60); repaint(); return (--counter); } return 1; } @Override public Dimension getPreferredSize() { return (new Dimension(500, 500)); } @Override public void paintComponent(Graphics g) { super.paintComponent(g); g.clearRect(0, 0, getWidth(), getHeight()); g.drawImage(image.getImage(), imageX, imageY, null); } private cl.

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