Answer:
(a) What is the best case time complexity of the algorithm (assuming n > 1)?
Answer: O(1)
(b) What is the worst case time complexity of the algorithm?
Answer: O(n^4)
Explanation:
(a) In the best case, the if condition will be true, the program will only run once and return so complexity of the algorithm is O(1)
.
(b) In the worst case, the program will run n^4 times so complexity of the algorithm is O(n^4).
A semiconductor material has an electrical conductivity value falling between that of a conductor, such as metallic copper, and an insulator, such as glass. Its resistivity falls as its temperature rises; metals behave in the opposite way. I hope this helps and have a great day! (Also brainliest would be appreciated but you don’t have to) :)
Answer:
1. True 2. False 3. True 4. True 5. True 6. False 7. True
Explanation:
1. A hacker/cracker finds and exploits weakness in order to gain access with a criminal intent, just as an intruder.
2. Activists are people who campaign to bring about a positive political or social change.
3. It is illegal to use a device as a packet sniffer to steal people's usernames and passwords.
4. This quite true, as there is a huge community of hackers where people are highly recognized for hacking.
5. Yes, intruders have a common attack methodology.
6. IDS monitors networks or systems to identify suspicious activities while a user interface is the means the computer and the user interacts. So it's false.
7. instrusion detection involves monitoring networks or systems to identify suspicious activities, so an intruder is detected if their behavior is suspicious when compared to a legitimate user.
Answer:
punctuation_chars = ["'", '"', ",", ".", "!", ":", ";", '#', '@']
def strip_punctuation(strWord):
for charPunct in punctuation_chars:
strWord = strWord.replace(charPunct, "")
return strWord
Explanation:
The function is defined with a single argument.
A for loop is ran to check each character of the the word.
If a punction mark is present as a character in the word, it is removed.
The same word is returned but without the punctuation marks.
