LeetCode 2047: Number of Valid Words in a Sentence Solution

Master LeetCode problem 2047 (Number of Valid Words in a Sentence), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2047. Number of Valid Words in a Sentence

Easy

String

Problem Explanation

Explanation:

To solve this problem, we need to iterate through each token in the given sentence and check if it meets the criteria to be a valid word. We can do this by applying the specified rules for a valid word: containing only lowercase letters, hyphens, and/or punctuation, having at most one hyphen surrounded by lowercase characters, and having at most one punctuation mark at the end.

We can implement the solution by splitting the sentence into tokens using spaces as delimiters. Then, for each token, we can check if it satisfies the conditions to be a valid word. If a token is a valid word, we increment a counter. Finally, we return the total count of valid words found in the sentence.

Time complexity: O(n), where n is the length of the input sentence.
Space complexity: O(1) as we are not using any additional data structures that grow with the input size.

Solution Code

class Solution {
    public int countValidWords(String sentence) {
        String[] tokens = sentence.split("\\s+");
        int count = 0;
        for (String token : tokens) {
            if (isValidWord(token)) {
                count++;
            }
        }
        return count;
    }

    private boolean isValidWord(String word) {
        if (word.isEmpty() || !Character.isLetter(word.charAt(0))) {
            return false;
        }
        boolean hasHyphen = false;
        boolean hasPunctuation = false;
        for (int i = 0; i < word.length(); i++) {
            char c = word.charAt(i);
            if (Character.isLetter(c)) {
                continue;
            } else if (c == '-') {
                if (hasHyphen || i == 0 || i == word.length() - 1 || !Character.isLetter(word.charAt(i - 1)) || !Character.isLetter(word.charAt(i + 1))) {
                    return false;
                }
                hasHyphen = true;
            } else if (i == word.length() - 1 && (c == '.' || c == '!' || c == ',')) {
                hasPunctuation = true;
            } else {
                return false;
            }
        }
        return !hasHyphen || hasPunctuation;
    }
}

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