192.168 suggests class C networks which have a 24 bit netmask (255.255.255.0) but you haven't provided enough info, like the netmask of a working machine, to be definite.
Answer:
Explanation:
The following code is written in Python. It asks the user for an input. Then cleans the input using regex to remove all commas, whitespace, and apostrophes as well as making it all lowercase. Then it reverses the phrase and saves it to a variable called reverse. Finally, it compares the two versions of the phrase, if they are equal it prints out that it is a palindrome, otherwise it prints that it is not a palindrome. The test case output can be seen in the attached picture below.
import re
phrase = input("Enter word or phrase: ")
phrase = re.sub("[,'\s]", '', phrase).lower()
reverse = phrase[::-1]
if phrase == reverse:
print("This word/phrase is a palindrome")
else:
print("This word/phrase is NOT a palindrome")
Answer:
See explaination for the program code
Explanation:
The code below
Pseudo-code:
//each item ai is used at most once
isSubsetSum(A[],n,t)//takes array of items of size n, and sum t
{
boolean subset[n+1][t+1];//creating a boolean mtraix
for i=1 to n+1
subset[i][1] = true; //initially setting all first column values as true
for i = 2 to t+1
subset[1][i] = false; //initialy setting all first row values as false
for i=2 to n
{
for j=2 to t
{
if(j<A[i-1])
subset[i][j] = subset[i-1][j];
if (j >= A[i-1])
subset[i][j] = subset[i-1][j] ||
subset[i - 1][j-set[i-1]];
}
}
//returns true if there is a subset with given sum t
//other wise returns false
return subset[n][t];
}
Recurrence relation:
T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times
T(1)=1
solving recurrence:
T(n)=T(n-1)+t
T(n)=T(n-2)+t+t
T(n)=T(n-2)+2t
T(n)=T(n-3)+3t
,,
,
T(n)=T(n-n-1)+(n-1)t
T(n)=T(1)+(n-1)t
T(n)=1+(n-1)t = O(nt)
//so complexity is :O(nt)//where n is number of element, t is given sum
If you mean a web page ten the bet option would be elements, where gives you an unordered list and give you an order list
