Answer:
class Car(object):
fuel = 0
def __init__(self, mpg):
self.mpg = mpg
def drive(self, mile):
if self.fuel * self.mpg >= mile:
self.fuel -= mile / self.mpg
else:
print(f"get gas for your {self}")
print(f"Fuel remaining: {self.fuel}")
#classmethod
def get_gas(cls):
cls.fuel += 50
#classmethod
def add_gas(cls, gallon):
if cls.fuel + gallon > 50:
cls.fuel += 10
else:
cls.fuel += gallon
gulf = Car(20)
gulf.get_gas()
gulf.drive(200)
Explanation:
The Car class is defined in Python. Its drive method simulates the driving of a car with fuel that reduces by the miles covered, with efficiency in miles per gallon. The get_gas and add_gas methods fill and top up the car tank respectively.
Answer:
In Java:
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String isbn;
System.out.print("First 1:2 digits: ");
isbn = input.nextLine();
if(isbn.length()==12){
int chksum = 0;
for(int i = 0; i<12;i++){
if((i+1)%2==0){ chksum+= 3 * Character.getNumericValue(isbn.charAt(i)); }
else{ chksum+=Character.getNumericValue(isbn.charAt(i)); } }
chksum%=10;
chksum=10-chksum;
if(chksum==10){
System.out.print("The ISBN-13 number is "+isbn+"0");}
else{
System.out.print("The ISBN-13 number is "+isbn+""+chksum); } }
else{
System.out.print("Invalid Input");
} }}
Explanation:
See attachment for explanation where comments are used to explain each line
Answer: You must have the right soffware have the right research and have the right website.
Explanation: You need the right website or you interview will be messed up.
Answer:
Explanation:
The following code was written in Java and performs the exact requirements listed in the question. It has also been tested by the runner code and works perfectly.
public static boolean insert(String[] words, String newWord, int place) {
if (place > words.length) {
return false;
} else {
for (int x = words.length - 1; x >= 0; x--) {
if (place == x) {
words[x] = newWord;
break;
} else {
words[x] = words[x-1];
}
}
return true;
}
}
Answer:
The answer to this question is O(NlogN).
Explanation:
The time complexity of Heap Sort on an array A is O(NLogN) even if the array is already sorted in increasing order.Since the Heap Sort is implemented by creating the heap from the array and then heapifying and then repeatedly swapping first and last element and deleting the last element.The process will be done for the whole array.So the running time complexity is O(NLogN).
