Editorial

Solution walkthrough

Review the explanation, then compare the reference implementation.

Video explanation

Solution

Intuition

A word is typeable exactly when none of its characters uses a broken key. Words are independent, and a single broken character is enough to reject a word immediately.

Approach

Split the text into its space-separated words and store the broken letters in a set. For each word, test its characters for membership in that set; skip the word on the first match and count it otherwise. Return the number of accepted words. Repeated occurrences of a word count separately because the task counts words in the text, not distinct spellings. With no broken letters, every word passes.

class Solution:
    def canBeTypedWords(self, text: str, brokenLetters: str) -> int:
        text = text.split(' ')
        brokenLetters = set(brokenLetters)
        res = 0

        for w in text:
            if any(c in brokenLetters for c in w):
                continue
            res += 1

        return res

Complexity analysis

Let c be the total characters in text and b = brokenLetters.length. Time: O(c + b). Auxiliary space: O(c + b) for the split words and broken-letter set.