PLEASE CODE IN PYTHON

Write a function find worst brand() that searches a list of cars and returns the brand of car that has the worst repair record in terms of total repair costs. Since there might be several cars of the same brand, it will be necessary to sum the repair costs for all cars of each brand, determine which brand had the greatest total repair cost over all cars in the list, and finally, return the name of that brand. If two (or more) brands have the same maximum total repair code, the function can return either brand.

Hint: consider using a dictionary, where the keys are brands and the values are the total costs for the brands.

Another hint: to retrieve a list of the keys in a dictionary, you can use this syntax:

my_dict = { }

...code that stores key/value pairs in my_dict...

key_list = my_dict.keys()

The idea then is to iterate over the list of keys, find the maximum value, and return the key for that maximum value. In the context of the find worst brand() function, finding the key with the largest value is equivalent to finding the brand with the highest total cost.

PROVIDED BELOW ARE THE CLASSES THAT ARE BEING REFERENCED AS WELL AS THE LISTS OF CARS THAT WILL BE USED IN THE PROBLEM. PLEASE CODE ACCORDING TO THE PROVIDED DETAILS

if __name__ == '__main__': def reset_car_database(): c01 = Car('XYZ123X', 'Toyota', 'Camry', 2012) c02 = Car('HSY113Y', 'Honda', 'Civic', 2016) c03 = Car('MZJ291E', 'Ford', 'Escape', 2009) c04 = Car('KJD922P', 'Jeep', 'Wrangler', 2011) c05 = Car('TRQ235K', 'Hyundai', 'Sonata', 2017) c06 = Car('JNH47GB', 'Toyota', 'Camry', 2011) c07 = Car('K83JDE3', 'Honda', 'Pilot', 2009) c08 = Car('MCJD83J', 'Hyundai', 'Elantra', 2013) c09 = Car('9EM2JSK', 'Toyota', 'Camry', 2002) c10 = Car('JF83JKS', 'Honda', 'Civic', 2012) c01.add_repair('Broken axle', 2900) c01.add_repair('Punctured tire', 40) c02.add_repair('Cracked windshield', 1000) c04.add_repair('Oil change', 45) c04.add_repair('New clearcoat', 550) c05.add_repair('Punctured tire', 30) c05.add_repair('Cracked windshield', 1000) c06.add_repair('Popped dents', 75) c06.add_repair('Broken headlight', 80) c06.add_repair('Broken taillight', 95) c07.add_repair('Rebuilt engine', 4880) c09.add_repair('Broken headlight', 80) c09.add_repair('Punctured tire', 80) c10.add_repair('Replaced windshield wipers', 125) car_list1 = [c02, c03, c05, c06, c07, c08, c09] car_list2 = [c01, c03, c04, c05, c06, c08, c09, c10] car_list3 = [c03, c04, c08] return car_list1, car_list2, car_list3 

 cars1, cars2, cars3 = reset_car_database() print('Testing find_worst_brand() with cars1: ' + str(find_worst_brand(cars1))) print('Testing find_worst_brand() with cars2: ' + str(find_worst_brand(cars2))) print('Testing find_worst_brand() with cars3: ' + str(find_worst_brand(cars3))) 

EXAMPLES OF FINAL OUTPUT:

find worst brand(cars1) "Honda"

find worst brand(cars2) "Toyota"

find worst brand(cars3) "Jeep"