Answered: Return the index of the first integer…

Return the index of the first integer in items that is less than its index, or -1 if no such integer exists in items. >>> small_index([2, 5, 7, 99, 6]) -1

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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#include using namespace std; struct ListNode { string data; ListNode *next; }; int main() { ListNode *p, *list; list = new ListNode; list->data = "New York"; p new ListNode; p->data = "Boston"; list->next = p; p->next = new ListNode; p->next->data = "Houston"; p->next->next = nullptr; // new code goes here Which of the following code correctly deletes the node with value "Boston" from the list when added at point of insertion indicated above? O list->next = p; delete p; O p = list->next; %3D list->next = p->next; delete p; p = list->next; list = p->next; delete p; O None of these O p = list->next; %3D list->next = p; %3D delete p;
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;…
В.width; } { t = B.type; w = { T.type = C.type; T.width = C.width; } T → B C В > int { B.type = integer; B.width 4; } В — foat { B.type = float; B.width = 8; } { C.type = t; C.width = w; } C - [ num] C1 = array(num. value, C1.type); { C.type C.width = num. value x C1. width; }
Count dominators def count_dominators(items): An element of items is said to be a dominator if every element to its right (not just the one element that is immediately to its right) is strictly smaller than it. By this definition, the last item of the list is automatically a dominator. This function should count how many elements in items are dominators, and return that count. For example, dominators of [42, 7, 12, 9, 13, 5] would be the elements 42, 13 and 5. Before starting to write code for this function, you should consult the parable of "Shlemiel the painter" and think how this seemingly silly tale from a simpler time relates to today's computational problems performed on lists, strings and other sequences. This problem will be the first of many that you will encounter during and after this course to illustrate the important principle of using only one loop to achieve in a tiny fraction of time the same end result that Shlemiel achieves with two nested loops. Your workload…
#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;…
#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…
Pointers: P3: Given the following string, after you call func2, if you printed out the array, what would be printed out? string a = {"c","a","n", func2(a); "y"}; void func2(string *arr) { arr[3) - "t"; arr[2]="vi"; %3D
#include<iostream> #include<cstdlib> #include<string> #include<cstdio> using namespace std; const int TABLE_SIZE = 128; /* * HashEntry Class Declaration */ class HashEntry { public: int key; int value; HashEntry(int key, int value) { this->key = key; this->value = value; } }; /* * HashMap Class Declaration */ class HashMap { private: HashEntry **table; public: HashMap() { table = new HashEntry * [TABLE_SIZE]; for (int i = 0; i< TABLE_SIZE; i++) { table[i] = NULL; } } /* * Hash Function */ int HashFunc(int key) { return key % TABLE_SIZE; } /* * Insert Element at a key */ void Insert(int key, int value) { int hash = HashFunc(key); while (table[hash] != NULL && table[hash]->key != key)…
#include<iostream> #include<cstdlib> #include<string> #include<cstdio> using namespace std; const int TABLE_SIZE = 128; /* * HashEntry Class Declaration */ class HashEntry { public: int key; int value; HashEntry(int key, int value) { this->key = key; this->value = value; } }; /* * HashMap Class Declaration */ class HashMap { private: HashEntry **table; public: HashMap() { table = new HashEntry * [TABLE_SIZE]; for (int i = 0; i< TABLE_SIZE; i++) { table[i] = NULL; } } /* * Hash Function */ int HashFunc(int key) { return key % TABLE_SIZE; } /* * Insert Element at a key */ void Insert(int key, int value) { int hash = HashFunc(key); while (table[hash] != NULL && table[hash]->key != key)…
ion C Lons Employee -id: int -name String - dob : Date -staff : ArrayList +setid(int): void +getld() : int +setName(String): void +setDob(Date) : void +getDob(): Date +addStaff(Employee) : void +getStaff() : ArrayList
Dec2Hex function : def decimal_to_hex(number): hex_map = {0: '0', 1: '1', 2: '2', 3: '3', 4: '4', 5: '5', 6: '6', 7: '7', 8: '8', 9: '9', 10: 'A', 11: 'B', 12: 'C', 13: 'D', 14: 'E', 15: 'F'} hex_digits = [] while number > 0: hex_digits.append(hex_map[number % 16]) number //= 16 # Reverse the order of the hex digits and join them to form the final hex string hex_string = ''.join(reversed(hex_digits)) return f'0x{hex_string}' if hex_string else '0x0' Perform a Profile of your Dec2Hex function. Write a function that takes a timedate object as a parameter andcalculates the number of years from NOW to the time in the timedateobject. Add unit-testing code to your Dec2Hex exercise, and then perform aUnit test of the updated Dec2Hex code.
#ifndef lab5ExF_h #define lab5ExF_h typedef struct point { char label[10]; double x ; // x coordinate for point in a Cartesian coordinate system double y; // y coordinate for point in a Cartesian coordinate system double z; // z coordinate for point in a Cartesian coordinate system }Point; void reverse (Point *a, int n); /* REQUIRES: Elements a[0] ... a[n-2], a[n-1] exists. * PROMISES: places the existing Point objects in array a, in reverse order. * The new a[0] value is the old a[n-1] value, the new a[1] is the * old a[n-2], etc. */ int search(const Point* struct_array, const char* target, int n); /* REQUIRES: Elements struct-array[0] ... struct_array[n-2], struct_array[n-1] * exists. target points to string to be searched for. * PROMISES: returns the index of the element in the array that contains an * instance of point with a matching label. Otherwise, if there is * no point in the array that its label matches the target-label, * it should return -1. * If there are more than…