from statistics import median, mean, mode

def print_stats(array):
  print(array)
  print("median =", median(array))
  print("mean   =", mean(array))
  print("mode   =", mode(array))
  print()

print_stats([1, 2, 3, 3, 4])
print_stats([1, 2, 3, 3])

import itertools
import string
import time

def guess_password(real):
    chars = string.ascii_lowercase + string.ascii_uppercase + string.digits + string.punctuation
    attempts = 0
    for password_length in range(1, 9):
        for guess in itertools.product(chars, repeat=password_length):
            startTime = time.time()
            attempts += 1
            guess = ''.join(guess)
            if guess == real:
                return 'password is {}. found in {} guesses.'.format(guess, attempts)
            loopTime = (time.time() - startTime);
            print(guess, attempts, loopTime)

print("\nIt will take A REALLY LONG TIME to crack a long password. Try this out with a 3 or 4 letter password and see how this program works.\n")
val = input("Enter a password you want to crack that is 9 characters or below: ")
print(guess_password(val.lower()))

from collections import defaultdict

def collect_dictionary(obj):
  inv_obj = defaultdict(list)
  for key, value in obj.items():
    inv_obj[value].append(key)
  return dict(inv_obj)
  
ages = {
  'Peter': 10,
  'Isabel': 10,
  'Anna': 9,
}
collect_dictionary(ages) # { 10: ['Peter', 'Isabel'], 9: ['Anna'] }

# Python binary search function
def binary_search(arr, target):
    left = 0
    right = len(arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

# Usage
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
target = 7
result = binary_search(arr, target)
if result != -1:
    print(f"Element is present at index {result}")
else:
    print("Element is not present in array")

print(123)

from collections import Counter

def find_parity_outliers(nums):
  return [
    x for x in nums
    if x % 2 != Counter([n % 2 for n in nums]).most_common()[0][0]
  ]

find_parity_outliers([1, 2, 3, 4, 6]) # [1, 3]