Answer:
Forward the packet to the next hop router specified by the route network 0.0.0.0
Explanation:
Answer:
- def check_subset(l1, l2):
- status = False
- count = 0
- if(len(l1) > len(l2)):
- for x in l2:
- for y in l1:
- if x == y:
- count += 1
-
- if(count == len(l2)):
- return True
- else:
- return False
-
- else:
- for x in l1:
- for y in l2:
- if x==y:
- count += 1
-
- if(count == len(l1)):
- return True
- else:
- return False
-
- print(check_subset([1,4,6], [1,2,3,4,5,6]))
- print(check_subset([2,5,7,9,8], [7,8]))
- print(check_subset([1, 5, 7], [1,4,6,78,12]))
Explanation:
The key idea of this solution is to create a count variable to track the number of the elements in a shorter list whose value can be found in another longer list.
Firstly, we need to check which list is shorter (Line 4). If the list 2 is shorter, we need to traverse through the list 2 in an outer loop (Line 5) and then create another inner loop to traverse through the longer list 1 (Line 6). If the current x value from list 2 is matched any value in list 1, increment the count variable by 1. After finishing the outer loop and inner loop, we shall be able to get the total count of elements in list 2 which can also be found in list 1. If the count is equal to the length of list 2, it means all elements in the list 2 are found in the list 1 and therefore it is a subset of list 1 and return true (Line 10-11) otherwise return false.
The similar process is applied to the situation where the list 1 is shorter than list 2 (Line 15-24)
If we test our function using three pairs of input lists (Line 26-28), we shall get the output as follows:
True
True
False
Answer:
public ArrayList onlyBlue(String[] clothes){
ArrayList<String> blueCloths = new ArrayList<>();
for(int i =0; i<clothes.length; i++){
if(clothes[i].equalsIgnoreCase("blue")){
blueCloths.add(clothes[i]);
}
}
return blueCloths;
}
Explanation:
- Create the method to accept an Array object of type String representing colors with a return type of an ArrayList
- Within the method body, create and initialize an Arraylist
- Use a for loop to iterate the Array of cloths.
- Use an if statement within the for loop to check if item equals blue and add to the Arraylist.
- Finally return the arrayList to the caller
Answer:
Basically, dealing with the "software crisis" is what we now call software engineering. We just see the field more clearly now.
What this crisis was all about is that in the early days of the modern technological era -- in the 1950s, say -- there was tremendous optimism about the effect that digital computers could have on society, on their ability to literally solve humanity's problems. We just needed to formalize important questions and let our hulking "digital brains" come up with the answers.
Artificial intelligence, for example, had some early successes in easy to formalize domains like chess and these sorts of successes led to lots of people who should have known better making extremely naive predictions about how soon perfect machine translation would transform human interaction and how soon rote and onerous work would be relegated to the dustbin of history by autonomous intelligent machines.
Answer:
#include<stdio.h>
//declare a named constant
#define MAX 50
int main()
{
//declare the array
int a[MAX],i;
//for loop to access the elements from user
for(i=0;i<MAX;i++)
{
printf("\n Enter a number to a[%d]",i+1);
scanf("%d",&a[i]);
}
//display the input elements
printf("\n The array elements are :");
for(i=0;i<=MAX;i++)
printf(" %d ",a[i]);
}
Explanation:
PSEUDOCODE INPUTARRAY(A[MAX])
REPEAT FOR I EQUALS TO 1 TO MAX
PRINT “Enter a number to A[I]”
READ A[I]
[END OF LOOP]
REPEAT FOR I EQUALS TO 1 TO MAX
PRINT A[I]
[END OF LOOP]
RETURN
ALGORITHM
ALGORITHM PRINTARRAY(A[MAX])
REPEAT FOR I<=1 TO MAX
PRINT “Enter a number”
INPUT A[I]
[END OF LOOP]
REPEAT FOR I<=1 TO MAX
PRINT A[I]
[END OF LOOP]
