✅ Simple Straightforward Solution ✅
IntuitionThe core idea is to efficiently check if each character in a word exists within the allowed characters. Using a set for allowed allows for quick lookup
# Intuition
The core idea is to efficiently check if each character in a word exists within the allowed characters. Using a set for allowed allows for quick lookups.

# Approach

1. Create a set: Convert the allowed string into a set (allowed_set) for efficient character lookups.
2. Iterate through words: Loop through each word in the words list.
3. Check for consistency:  For each word:
    - Use all(ch in allowed_set for ch in word) to check if every character (ch) in the word is present in the allowed_set. The all() function efficiently checks this condition for all characters.
    - If all characters are in allowed_set (meaning the word is consistent), increment the counter c.
4. Return the count: After checking all words, return the final count c of consistent strings.

# Complexity
- Time complexity: $$O(m * n) $$
where n is the total number of characters in all words in the words list, and m is the average length of a word in words. This is because we iterate through each character of each word in the words list.  The ch in allowed_set check takes approximately constant time due to the use of a set.

- Space complexity: $$O(k) $$
where k is the number of unique characters in the allowed string. This is the space used to store the allowed_set. Since allowed only contains distinct characters, k will be at most 26 (for lowercase alphabets).

# Code
```python3 []
class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        allowed_set = set(allowed)
        c = 0
        for word in words:
            if all(ch in allowed_set for ch in word):
                c += 1
        return c

```
['Array', 'Python3']