Implement most_frequent_key without using max()

Q: Implement most_frequent_key without using max()

This question evaluates proficiency in Python fundamentals and object-oriented inheritance, focusing on frequency aggregation using a precomputed dictionary, iteration over mappings, and correct handling of ties and empty inputs.

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Question

Problem (Python OOP)

You are given two classes. Parent precomputes frequency counts of items (as strings) from an input list.

class Parent:
    def __init__(self, l):
        self.l = l
        # set of all unique string forms
        self.lf = set(str(s) for s in self.l)

        # dict: key -> count
        self.ldf = {}
        for item in self.l:
            s = str(item)
            if s not in self.ldf:
                self.ldf[s] = 1
            else:
                self.ldf[s] += 1


class Child(Parent):
    def most_frequent_key(self):
        """Return the key(s) with the highest frequency.

        Constraints:
        - You may NOT use Python's built-in `max()`.
        - Use the precomputed `self.ldf` from `Parent`.

        Return:
        - A list of the most frequent key(s) (strings). If there is a tie, include all tied keys.
        - If the input list is empty, return an empty list.
        """
        # TODO: implement
        return output_list

Example

queries = ['park', 'Park', 'McDonalds', 'apple', 'Apple Park', 'park', 'Park']

x = Child(queries)
print("most frequent:", x.most_frequent_key())

Notes/clarifications

Counting is case-sensitive because keys are created via str(item) .
The goal is to manually compute the highest frequency via iteration (no max() ).

Implement most_frequent_key without using max()

Overview

Problem (Python OOP)

Example

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