Solution: count characters

text = """
This is a very long text.
OK, maybe it is not that long after all.
"""

# print(text)
count = {}

for char in text:
    if char == '\n':
        continue
    if char not in count:
        count[char] = 1
    else:
        count[char] += 1

for key in sorted( count.keys() ):
    print("'{}' {}".format(key, count[key]))

• We need to store the counter somewhere. We could use two lists for that, but that would give a complex solution that runs in O(n**2) time.
• Besides, we are in the chapter about dictionaries so probably we better use a dictionary.
• In the count dictionary we each key is going to be one of the characters and the respective value will be the number of times it appeared.
• So if out string is "aabx" then we'll end up with

{
    "a": 2,
    "b": 1,
    "x": 1,
}

• The for in loop on a string will iterate over it character by charter (even if we don't call our variable char.

• We check if the current character is a newline \n and if it we call continue to skip the rest of the iteration. We don't want to count newlines.

• Then we check if we have already seen this character. That is, it is already one of the keys in the count dictionary. If not yet, then we add it and put 1 as the values. After all we saw one copy of this character. If we have already seen this character (we get to the else part) then we increment the counter for this character.

• We are done now with the data collection.

• In the second loop we go over the keys of the dictionary, that is the characters we have encountered. We sort them in ASCII order.

• Then we print each one of them and the respective value, the number of times the character was found.