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# Teacher Stat Challenge Task 4 of 5.... Need help getting number of teachers and comparing them

Alright for my code, I need help with the most_courses function. This is where I added a skeleton structure but will need help figuring out how to compare multiple teachers instead of just 2. Any suggestions/help?

teachers.py
```# The dictionary will look something like:
# {'Andrew Chalkley': ['jQuery Basics', 'Node.js Basics'],
#  'Kenneth Love': ['Python Basics', 'Python Collections']}
#
# Each key will be a Teacher and the value will be a list of courses.
#
# Your code goes below here.
def num_teachers(arg):
return len(arg)
def num_courses(arg):
return sum(len(v) for v in arg.values())
def courses(teachers):
# create empty list named 'stuff'
stuff = []
# extend the values into the empty list--now it should have 2 lists in the list
stuff.extend(teachers.values())
# Sketchy/Questionable part: Created a variable with the number of values that was in the dictionary
# Need to figure out how to loop that as many times as the values and add it to the list
numbers = len(teachers)
index = 0
stuffed = []
while index != numbers:
stuffed += stuff[index]
index += 1
return stuffed

def most_courses(teachers):
# Create empty list
stuffed3 = []
# Append the keys to the empty list
stuffed3.extend(teachers.keys())
# Sketchy/Questionable part: Return teacher name with the most courses
# Get number of teachers, subtracting 1 to take into account index starting at 0
numbers = len(teachers) - 1
index = 0
while index != numbers:
if len(stuffed3[0]) > len(stuffed3[1]):
return stuffed3[0]
else:
return stuffed3[1]
```

This solution will solve the challenge

```def most_courses(arg):
max_count = 0
teachers = " "
for teacher, listOfCourses in arg.items():
if len(listOfCourses) > (max_count):
max_count = len(listOfCourses)
teachers = teacher
return teachers
```

This function will return the first teacher in the dict with the most courses, and will not take into consideration that there can be more teachers that have the same amount of courses. But it will give you a pass on the challenge.