2 Consider the sequence of keys (5,16,22,45,2,10,18,30,50,12,1). Draw the result of inserting entries with these keys (in the gi
Juliette [100K]
Answer:
A) (2,4) tree
- Insertion of key 45 makes key unbalanced and this is because it violates the 2,4 tree so we split the node
- insertion of key 10 makes key unbalanced and this is because it violates the 2,4 tree so we split the node
B) Red-black tree
Explanation:
The diagrams for the solutions are attached showing the results of inserting entries
Answer:
import java.util.Scanner;
public class TestClock {
public static void main(String[] args) {
Scanner in = new Scanner (System.in);
System.out.print("Enter favorite color:");
String word1 = in.next();
System.out.print("Enter pet's name:");
String word2 = in.next();
System.out.print("Enter a number:");
int num = in.nextInt();
System.out.println("you entered: "+word1+" "+word2+" "+num);
}
}
Explanation:
Using Java Programming language
- Import the Scanner class
- create an object of the scanner class
- Prompt user to enter the values for the variables (word1, word2, num)
- Use String concatenation in System.out.println to display the output as required by the question.
Answer:
C++ code is given below
Explanation:
#include <iostream>
#include <cctype>
#include <string.h>
#include <cstring>
#include <sstream>
using namespace std;
struct Car {
public:
char reportingMark[5];
int carNumber;
string kind;
bool loaded;
string destination;
};
void input(Car *);
void output(Car *);
int main() {
Car *T = new Car;
input(T);
output(T);
delete T;
return 0;
}
void input(Car *T)
{
string str, s;
cout << " Enter the reporting mark as a 5 or less character uppercase string: ";
cin >> str;
for (int i = 0; i < str.length(); i++)
T->reportingMark[i] = toupper(str[i]);
cout << " Enter the car number: ";
cin >> T->carNumber;
cout << " Enter the kind: ";
cin >> T->kind;
cout << " Enter the loaded status as true or false: ";
cin >> s;
istringstream(s) >> boolalpha >> T->loaded;
if (T->loaded == true) {
cout << " Enter the destination: ";
cin.ignore();
getline(cin, T->destination);
}
else
T->destination = "NONE";
}
void output(Car *T)
{
cout << " Reporting Mark: " << T->reportingMark;
cout << " Car Number: " << T->carNumber;
cout << " Kind: " << T->kind;
cout << " Loaded Status: " << boolalpha << T->loaded;
cout << " Destination: " << T->destination << " ";
}
Aaron's action is considered legal, as you are allowed to make a backup copy of a legal copy of a software, but it can only be used in case the original software is destroyed or unusable.
Answer:
The answer with explanation is attached in the file.
Explanation:
