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.
The answer is C) tablet. You use the touch screen keyboard for that.
Answer:
retupmoc
Explanation:
1.) Anwser will be retupmoc
because
public static String mysteryString(String s){
if(s.length() == 1){
return s;
}
else{
return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));
}
}
In this program input is "computer" . So the function mysteryString(String s) it does
return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));
so when it enters the first time ??s.substring(s.length() -1) and it will be give you 'r' then it calls the function recursively by reducing the string length by one . So next time it calls the mysteryString function with string "compute" and next time it calls return s.substring(s.length()-1)? + mysteryString(s.substring(0,s.length-1)) so this time it gives "e" and calls the function again recursively . It keeps on doing till it matched the base case.
so it returns "retupmoc".
