1) Get users who attended some activity more than twice.

>> SELECT * FROM users;
user_id  username
1        John Doe
2        Jane Don
3        Alice Jones
4        Lisa Romero


>> SELECT * FROM training_details;
user_training_id  user_id  training_id  training_date
1                 1        1            "2015-08-02"
2                 2        1            "2015-08-03"
3                 3        2            "2015-08-02"
4                 4        2            "2015-08-04"
5                 2        2            "2015-08-03"
6                 1        1            "2015-08-02"
7                 3        2            "2015-08-04"
8                 4        3            "2015-08-03"
9                 1        4            "2015-08-03"
10                3        1            "2015-08-02"
11                4        2            "2015-08-04"
12                3        2            "2015-08-02"
13                1        1            "2015-08-02"
14                4        3            "2015-08-03"

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SELECT user.username, count(*)
FROM users
JOIN training_details as td ON td.user_id=users.user_idf
GROUP BY td.user_id

2) Implement a group_by_owners that:
Accepts a dictionary containing the file owner name for each file name.
Returns a dictionary containing a list of file names for each owner name, in any order.
For example, dictionary:
{
    'Input.txt': 'Randy',
    'Code.py': 'Stan',
    'Output.txt': 'Randy'
}
​
the group_by_owners function should return:
{
    'Randy': ['Input.txt', 'Output.txt'],
    'Stan': ['Code.py']
}

--------------------------------------------

def group_by_owners(file_owner_name_dict):
    result = {}
    users = set(file_owner_name_dict.values())
    for user in users:
        filelist = []
        for key in file_owner_name_dict:
            if file_owner_name_dict[key] == user:
                filelist.append(key)
        result.update({user: filelist})
    return result

    # for file, user in file_owner_name_dict.items():
    #    ...

3) Sort list of objects by age field in ascending order


Input:
animals = [
    {'type': 'penguin', 'name': 'Stephanie', 'age': 8},
    {'type': 'elephant', 'name': 'Devon', 'age': 3},
    {'type': 'puma', 'name': 'Moe', 'age': 5},
]

Output:
sorted_animals = [
    {'type': 'elephant', 'name': 'Devon', 'age': 3},
    {'type': 'puma', 'name': 'Moe', 'age': 5},
    {'type': 'penguin', 'name': 'Stephanie', 'age': 8}
]
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4) What will be the output of the following code snippet?

class Person:
    def __init__(self, id):
        self.id = id

sam = Person(100)
sam.__dict__['age'] = 49
print(sam.age + len(sam.__dict__))
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5) What will be the output of the following code snippet?

class A:
    def __init__(self):
        self.calcI(30)
        print("i from A is", self.i)
        
    def calcI(self, i):
        self.i = 2 * i
        
class B(A):
    def calcI(self, i):
        self.i = 3 * i
​
b = B()


class Animal:
    def x():
        self.say()

    def say():
        print('???')

class Cat(Animal):
    def say():
        print('moew')

class Dog(Animal):
    def say():
        print('wow')
    
Cat().x()

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6) Write a code snippet that returns all pairs of numbers that return sum of 10
[0, -1, 9, 9, 1, 3, 6, 7, 1]
--------------------------------------------