This is a java program with a class called Sum1.
Explanation:
While loop is used to add 10 number.
import java.util.Scanner;
public class Sum 1
{
public static void main(String args[])
{
int number, count, sum = 0;
Scanner s = new Scanner(System.in);
System.out.print("Enter the number:");
number = s.nextInt();
while(number > 0)
{
count = number % 10;
sum = sum + n;
number = number / 10;
}
System.out.println("Sum of Digits:"+sum);
}
}
Text = “ I really like owls. Did you know that an owls eyes are more than twice as big as the eyes of other birds of comparable weight? And that when an owl partially closes its eyes during the day, it is just blocking out light? Sometimes I wish I could be an owl.
word = ‘owl’
texts = text.lower()
owlist = list(texts.split())
count = text.count(word)
num = [owlist, count] #num has no meaning just random var
print(num)
Alter in anyway you want so that you can succeed. ✌
Answer:
I say when u don't know the answer completly.
Answer:
Finding kth element is more efficient in a doubly-linked list when compared to a singly-linked list
Explanation:
Assuming that both lists have firs_t and last_ pointers.
For a singly-linked list ; when locating a kth element, you have iterate through a number of k-1 elements which means that locating an element will be done only in one ( 1 ) direction
For a Doubly-linked list : To locate the Kth element can be done from two ( directions ) i.e. if the Kth element can found either by traversing the number of elements before it or after it . This makes finding the Kth element faster because the shortest route can be taken.
<em>Finding kth element is more efficient in a doubly-linked list when compared to a singly-linked list </em>
