Answer:
See explaination for the code
Explanation:
def wordsOfFrequency(words, freq):
d = {}
res = []
for i in range(len(words)):
if(words[i].lower() in d):
d[words[i].lower()] = d[words[i].lower()] + 1
else:
d[words[i].lower()] = 1
for word in words:
if d[word.lower()]==freq:
res.append(word)
return res
Note:
First a dictionary is created to keep the count of the lowercase form of each word.
Then, using another for loop, each word count is matched with the freq, if it matches, the word is appended to the result list res.
Finally the res list is appended.
<span>True
</span><span>There is an increasing need for private security personnel because of declining resources for many government agencies and a shortage of police officers in many communities due to lack of funding.
</span>
Answer:
more /var/log/auth.log
less /var/log/auth.log
Explanation:
The commands that can be used to view the content of the auth.log file page by page are:
more /var/log/auth.log
less /var/log/auth.log
