2047. Number of Valid Words in a Sentence

Description

A sentence consists of lowercase letters ('a' to 'z'), digits ('0' to '9'), hyphens ('-'), punctuation marks ('!', '.', and ','), and spaces (' ') only. Each sentence can be broken down into one or more tokens separated by one or more spaces ' '.

A token is a valid word if all three of the following are true:

  • It only contains lowercase letters, hyphens, and/or punctuation (no digits).
  • There is at most one hyphen '-'. If present, it must be surrounded by lowercase characters ("a-b" is valid, but "-ab" and "ab-" are not valid).
  • There is at most one punctuation mark. If present, it must be at the end of the token ("ab,", "cd!", and "." are valid, but "a!b" and "c.," are not valid).

Examples of valid words include "a-b.", "afad", "ba-c", "a!", and "!".

Given a string sentence, return the number of valid words in sentence.

 

Example 1:

Input: sentence = "cat and  dog"
Output: 3
Explanation: The valid words in the sentence are "cat", "and", and "dog".

Example 2:

Input: sentence = "!this  1-s b8d!"
Output: 0
Explanation: There are no valid words in the sentence.
"!this" is invalid because it starts with a punctuation mark.
"1-s" and "b8d" are invalid because they contain digits.

Example 3:

Input: sentence = "alice and  bob are playing stone-game10"
Output: 5
Explanation: The valid words in the sentence are "alice", "and", "bob", "are", and "playing".
"stone-game10" is invalid because it contains digits.

 

Constraints:

  • 1 <= sentence.length <= 1000
  • sentence only contains lowercase English letters, digits, ' ', '-', '!', '.', and ','.
  • There will be at least 1 token.

Solutions

Solution 1

Python Code
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class Solution:
    def countValidWords(self, sentence: str) -> int:
        def check(token):
            hyphen = False
            for i, c in enumerate(token):
                if c.isdigit() or (c in '!.,' and i < len(token) - 1):
                    return False
                if c == '-':
                    if (
                        hyphen
                        or i == 0
                        or i == len(token) - 1
                        or not token[i - 1].islower()
                        or not token[i + 1].islower()
                    ):
                        return False
                    hyphen = True
            return True

        return sum(check(token) for token in sentence.split())

Java Code
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class Solution {
    public int countValidWords(String sentence) {
        int ans = 0;
        for (String token : sentence.split(" ")) {
            if (check(token)) {
                ++ans;
            }
        }
        return ans;
    }

    private boolean check(String token) {
        int n = token.length();
        if (n == 0) {
            return false;
        }
        boolean hyphen = false;
        for (int i = 0; i < n; ++i) {
            char c = token.charAt(i);
            if (Character.isDigit(c) || (i < n - 1 && (c == '!' || c == '.' || c == ','))) {
                return false;
            }
            if (c == '-') {
                if (hyphen || i == 0 || i == n - 1 || !Character.isLetter(token.charAt(i - 1))
                    || !Character.isLetter(token.charAt(i + 1))) {
                    return false;
                }
                hyphen = true;
            }
        }
        return true;
    }
}

TypeScript Code
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function countValidWords(sentence: string): number {
    let words = sentence.trim().split(/\s+/);
    let ans = 0;
    for (let word of words) {
        if (isValied(word)) {
            ans++;
        }
    }
    return ans;
}

function isValied(str: string): boolean {
    let n = str.length;
    let hasLine = false;
    for (let i = 0; i < n; i++) {
        const char = str.charAt(i);
        if (/^[0-9]$/.test(char)) {
            return false;
        }
        if (char == '-') {
            if (hasLine) return false;
            else {
                hasLine = true;
            }
            let pre = str.charAt(i - 1),
                post = str.charAt(i + 1);
            if (!/^[a-z]$/g.test(pre) || !/^[a-z]$/g.test(post)) {
                return false;
            }
        }
        if (/^[\!\.\,\s]$/.test(char) && i != n - 1) {
            return false;
        }
    }
    return true;
}