count the number of consistent strings leetcode 1684 python

certainly! the problem “count the number of consistent strings” is a common exercise on leetcode (problem 1684). the objective is to determine how many strings from a given list are "consistent" with a set of allowed characters.

problem statement

you are given:
1. a string `allowed` consisting of distinct lowercase english letters.
2. an array of strings `words`.

a string is considered "consistent" if every character in the string exists in the `allowed` string. your task is to count how many strings from `words` are consistent.

steps to solve the problem

1. **convert allowed characters to a set**: this allows for o(1) average time complexity for lookups to check if a character is allowed.
2. **iterate through words**: for each word in the `words` array, check if all its characters are in the `allowed` set.
3. **count consistent strings**: maintain a counter that increments every time a consistent string is found.

python code example

here’s how you can implement this in python:

explanation of the code

- **line 1**: the function `countconsistentstrings` is defined, taking `allowed` and `words` as parameters.
- **line 3**: we convert the `allowed` string into a set called `allowed_set`. this will allow for fast membership testing.
- **line 6**: we initialize `consistent_count` to zero.
- **line 9**: we loop through each word in the `words` list.
- **line 11**: the `all()` function checks if all characters in the word are in `allowed_set`. if true, we increment the `consistent_count`.
- **line 13**: finally, we return the total count of consistent strings.

complexity analysis

- **time complexity**: o(n * m), where n is the number of words and m is the maximum length of the words. this is because we may need to check each character of each word.
- **space complexity**: o(k), where k is the size of the `allowed` set. this is the space used for storing the allowed characters.

conclusion

this approach efficiently counts the number of consistent strings ...

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