#1. There is a list of mac addresses - create a list of commands to delete this mac
#[‘aabbcc’, ‘a1b1c1’] -> [‘curl -X DELETE https://marina.ring.com/aabbcc’, ‘curl -X DELETE https://marina.ring.com/a1b1c1’]
macs = ['aabbcc', 'a1b1c1', 'cc1aa1']
commands = []

for n in macs:
    new_value=f'curl -X DELETE https://marina.ring.com/{n}'
    commands.append(new_value)
print(commands)

#2. There is a string, for each letter return the position of last occurrence.
#‘abracadabra’  -> {'a': 10, 'b': 8, 'r': 9, 'c': 4, 'd': 6}

letters="abracadabra"
result = {}
count = 0
for l in letters:
    result[l] = count
    count += 1
print(result)

#2.*There is a list of integers, create a dict with count of each element (have a look at defaultdict)
#[1, 2, 4, 6, 2, 7, 1, 2, 1, 2, 2] -> {1: 3, 2:5, 4:1, 6:1, 7:1}

from collections import defaultdict

int_list = [1, 2, 4, 6, 2, 7, 1, 2, 1, 2, 2]
dict1 = defaultdict(lambda: 0)

for m in int_list:
    dict1[m] += 1
print(dict(dict1))

#3. There is a list of integers, return count of even elements:
#[1, 2, 3, 4, 5, 6, 6, 6,7, 8, 22] -> 7
int_list = [1, 2, 3, 4, 5, 6, 6, 6,7, 8, 22]
count = 0
for n in int_list:
    if n%2==0:
        count += 1
print(count)

#4. There is a list of integers, return only those elements that can be divided by position index (add 0s)
#[0, 3, 3, 6, 12, 7, 4, 21] -> [0, 3, 6,12, 21]

ints = [0, 3, 3, 6, 12, 7, 4, 21]
result = []
started = False

for count, item in enumerate(ints):
    if not started:
        started = True
        result.append(item)
        continue
    if item%count==0:
        result.append(item)

print(result)

#5. Turn letters on even positions uppercase and on odd positions lowercase
#"quick BROWN fox JUMPS over THE lazy DOG" -> 'QuIcK BrOwN FoX JuMpS OvEr tHe lAzY DoG'

phrase = "quick BROWN fox JUMPS over THE lazy DOG"
new_phrase = []
count = 0
for item in phrase:
    if count%2==0:
        item = item.upper()
    new_phrase.append(item)
    count += 1
print(''.join(new_phrase))