Question

Can u guys help me solve this problem by using Python

Write function q1(listOfLists, infoDict). listOfLists is a list of zero or more lists of zero or more numbers. infoDict is a dictionary with numbers as keys. The values associated with keys can be of any type.

A number is considered "red" if the number is a key in the dictionary and has value "red". A number is considered "blue" if it is a key in the dictionary and has value "blue". Other numbers, whether or not they exist as keys in the dictionary, are considered "green".

A sublist of listOfLists is considered "red" if contains more red items than blue ones and more red items than green ones. A sublist is considered "blue" if it contains more blue items than red ones and more blue items than green ones. Other sublists are considered "green".

q1 returns a list that is the same length as listOfLists and such that item i is

• the largest element in listOfLists[i] if listOfLists[i] is red,
• the smallest element in listOfLists[i] if listOfLists[i] is blue,
• the most common element in listOfLists[i] if listOfLists[i] is green (if there is a tie for most common, take the smallest one, and use None if there are no elements)

For example

>>> q1([[4], [2,3,3], [1,2,3], [17]], {1:"purple", 2:"red", 3:"blue", 25:"red"})
[4, 2, 1, 17]

Answer 1

Your given snippet is in python, so I am writing in python

Code-

import collections

def q1(listOfLists:list, infoDict:dict):

    # result array

    res=[]

    # loop over all lists in listOfLists

    for lst in listOfLists:

        # variable to count the reds and blues

        red = 0

        blue = 0

        for ele in lst:

            if infoDict.get(ele, "") == 'red':

                red+=1

            elif infoDict.get(ele, "") == 'blue':

                blue +=1

        if red > blue:

            # If more red, insert larget element to list

            res.append(max(lst))

        elif blue > red:

            # If more blue, insert smallest element

            res.append(min(lst))

        else:

            # else, insert frequent element

            # counter element group same elements with their frequencies

            counter = collections.Counter(lst)

            # get most common element

            res.append(counter.most_common(1)[0][0])

    return res

print(q1([[4], [2,3,3], [1,2,3], [17]], {1:"purple", 2:"red", 3:"blue", 25:"red"}))

OUTPUT-

I hope it helps. For any doubt, feel free to ask in comments, and give upvote if u get the answer.

Edit:

Else block replacement-

else:

            # else, insert frequent element

            # counter element group same elements with their frequencies

            counter={}

            for ele in lst:

                #add elements to list to store their frequencies

                count[ele] = counter.get(ele, 0)+1

            #variable to store most frequent value

            mx_key=-1

            for key, value in counter.items():

                if value> counter.get(mx_key, 0):

                    mx_key = key

            res.append(mx_key)