B)C)int maxIndex = 0;for (int k = 1; k <= words.length; k++){if (words [k].compareTo (words [maxIndex]) > 0){}}return words [maxIndex];int maxIndex = 0;for (int k = 1; k< words.length; k++){if (words [k].compareTo (words [maxIndex]) > 0){}maxIndex = k;}return maxIndex;D) String maxWord = words [0];}maxIndex = k;for (int k = 1; k < words.length; k++){if (words [k].compareTo (maxWord) > 0){}maxWord k;}return maxWord;E) String maxWord = words [0];for (int k = 1; k< words.length; k++){if (words [k].compareTo(maxWord) > 0){maxWord words [k];}return maxword; dConsider the following instance variable and incomplete method. The method is intended to returna string from the array words that would be last alphabetically.private String [] words;public String findLastWord (){}/* missing implementation */Assume that words has been initialized with one or more strings containing only lowercase letters.Which of the following code segments can be used to replace / * missing implementation * / sothat findLastword will work as intended?A)int maxIndex = 0;for (int k = 0; k< words.length; k++){if (words [k].compareTo (max Index) > 0){}maxIndex k;}return words [maxIndex];

Answered: int maxIndex = 0; for (int k = 1; k <=…

int maxIndex = 0; for (int k = 1; k <= words.length; k++) { if (words [k].compareTo (words [maxIndex]) > 0) { } maxIndex = = k;

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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For any element in keysList with a value greater than 50, print the corresponding value in itemsList, followed by a comma (no spaces). Ex: If the input is: 32 105 101 35 10 20 30 40 the output is: 20,30, 1 #include 2 3 int main(void) { const int SIZE_LIST = 4; int keysList[SIZE_LIST]; int itemsList[SIZE_LIST]; int i; 4 6 7 8 scanf("%d", &keysList[0]); scanf ("%d", &keysList[1]); scanf("%d", &keysList[2]); scanf("%d", &keysList[3]); 10 11 12 13 scanf ("%d", &itemsList[0]); scanf ("%d", &itemsList[1]); scanf("%d", &itemsList[2]); scanf ("%d", &itemsList[3]); 14 15 16 17 18 19 /* Your code goes here */ 20 21 printf("\n"); 22 23 return 0; 24 }
Practice / Frequency of Characters Write a method that returns the frequency of each characters of a given String parameter. If the given String is null, then return null If the given String is empty return an empty map Example input: output: empty map explanation: input is empty Example input: null output: null explanation: input is null. Since problem output is focused on the frequency we can comfortably use Map data structure. Because we can use characters as key and the occurrences of them as value. Example input: responsible output: {r=1, e=2, s=2, p=1, o=1, n=1, i=1, b=1, l=1} explanation: characters are keys and oc values ences are
#include <stdio.h>#include <stdlib.h>#include <time.h> struct treeNode { struct treeNode* leftPtr; int data; struct treeNode* rightPtr;}; typedef struct treeNode TreeNode;typedef TreeNode* TreeNodePtr; void insertNode(TreeNodePtr* treePtr, int value);void inOrder(TreeNodePtr treePtr);void preOrder(TreeNodePtr treePtr);void postOrder(TreeNodePtr treePtr); int main(void) { TreeNodePtr rootPtr = NULL; srand(time(NULL)); puts("The numbers being placed in the tree are:"); for (unsigned int i = 1; i <= 10; ++i) { int item = rand() % 15; printf("%3d", item); insertNode(&rootPtr, item); } puts("\n\nThe preOrder traversal is: "); preOrder(rootPtr); puts("\n\nThe inOrder traversal is: "); inOrder(rootPtr); puts("\n\nThe postOrder traversal is: "); postOrder(rootPtr);} void insertNode(TreeNodePtr* treePtr, int value) { if (*treePtr == NULL) { *treePtr = malloc(sizeof(TreeNode)); if…
#include <stdio.h>#include <stdlib.h>#include <time.h> struct treeNode { struct treeNode *leftPtr; int data; struct treeNode *rightPtr;}; typedef struct treeNode TreeNode;typedef TreeNode *TreeNodePtr; void insertNode(TreeNodePtr *treePtr, int value);void inOrder(TreeNodePtr treePtr);void preOrder(TreeNodePtr treePtr);void postOrder(TreeNodePtr treePtr); int main(void) { TreeNodePtr rootPtr = NULL; srand(time(NULL)); puts("The numbers being placed in the tree are:"); for (unsigned int i = 1; i <= 10; ++i) { int item = rand() % 15; printf("%3d", item); insertNode(&rootPtr, item); } puts("\n\nThe preOrder traversal is: "); preOrder(rootPtr); puts("\n\nThe inOrder traversal is: "); inOrder(rootPtr); puts("\n\nThe postOrder traversal is: "); postOrder(rootPtr);} void insertNode(TreeNodePtr *treePtr, int value) { if (*treePtr == NULL) { *treePtr = malloc(sizeof(TreeNode)); if (*treePtr != NULL) { (*treePtr)->data = value;…
What input or parameter value impacts the number of times the recursive function will be called. • Give three specific examples of input/parameter values and, for each, state the number of times the recursive function will be called. • Devise a formula with respect to n that describes the number of times the recursive function will be called, where n is either the value passed or some property of the value passed (e.g. n might be the length of a string of the size of an array).
using namespace std; int main() int i, m=0,n=4; float arr[100] (55,66,88,1); for(i=0; i
def pairs(4): for ingredient in d: if ingredient[e] in d[ingredient]: return (ingredient, d[ingredient]) food - ('pasta': "parmesan, 'eges': " cheddar, 'pizza': 'mozzarella') print(pairs(food)) O (pasta. parmesan') O C'eggs: 'cheddar) O (pizza': 'mozzarella') O lpasta', parmesan), ('eggs': "cheddar)) O none of the above
void ExtractMinimumWeightPath(int path[MAX], int weights[MAX][MAX], int pathWeights[MAX][MAX], int rowc, int colc){int rowsCount = rowc;int colsCount = colc;path[rowsCount];int col = 0;int i = 0;for(i=1;i<colsCount;i++){if(pathWeights[rowsCount-1][i]<pathWeights[rowsCount-1][col]){col = i;}} int row = rowsCount-1;do{path[row] = col + 1;if (col>0 && pathWeights[row-1][col-1] + weights[row][col] == pathWeights[row][col]){col = col - 1;} else if (col < colsCount-1 && pathWeights[row-1][col+1] + weights[row][col] == pathWeights[row][col]){col = col + 1;}row--;} while(row>0);path[0] = col + 1;} This is a code in C, I need to change the code as much as possible, but on condition to perform the same work as this. Changing the var names and the loops is enough
public class Program7 { publicstaticvoidmain(String[]args){ int[]numbers=newint[100]; String[]names={"Maria","Aris","Galin","Galena"}; double[]balances=newdouble[]{2,56,12.57,36.57,57.89}; //length System.out.println(numbers.length); System.out.println(names.length); System.out.println(balances.length); //modifying names[3]="Celine Dion"; System.out.println("__________________________"); //looping for(inti=0;i<names.length;i++){ System.out.println(names[i]); } System.out.println("__________________________"); for(Stringname:names){ System.out.println(name); } //modifying array System.out.println("__________________________"); for(inti=0;i<names.length;i++){ names[i]=names[i]+" Annan"; System.out.println(names[i]); } System.out.println("__________________________"); doubletotal=0.0; for(inti=0;i<balances.length;i++){ total+=balances[i]; } System.out.println("Total Balance: "+ total); } } //calculate the average of the total balance
@return index of the point that is closest to the origin, which is (0, 0) * In case of a tie, return the lowest index */ public int closestToOriginIndex() { return 0; }
Q: Convert this to sorted array #include<iostream> #include"Student.cpp" class StudentList { private: struct ListNode { Student astudent; ListNode *next; }; ListNode *head; public: StudentList(); ~StudentList(); int IsEmpty(); void Add(Student newstudent); void Remove(); void DisplayList(); }; StudentList::StudentList() { head=NULL; }; StudentList::~StudentList() { cout <<"\nDestructing the objects..\n"; while(IsEmpty()!=0) Remove(); if(IsEmpty()==0) cout <<"All students have been deleted from a list\n"; }; int StudentList::IsEmpty() { if(head==NULL) return 0; else return 1; }; void StudentList::Add(Student newstudent) { ListNode *newPtr=new ListNode; if(newPtr==NULL) cout <<"Cannot allocate memory"; else { newPtr->astudent=newstudent; newPtr->next=head; head=newPtr; } }; void StudentList::Remove() { if(IsEmpty()==0) cout <<"List empty on remove"; else { ListNode *temp=head;…
def bigoh2(n: int) -> int: res = 0 for i in range(n): for j in range(i, n // 2): res += (i * j) return res