LeetCode #2047 — EASY

Number of Valid Words in a Sentence

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"cat and  dog"

Example 2

"!this  1-s b8d!"

Example 3

"alice and  bob are playing stone-game10"

Core Insight

What unlocks the optimal approach

Iterate through the string to split it by spaces.
Count the number of characters of each type (letters, numbers, hyphens, and punctuations).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2047: Number of Valid Words in a Sentence
class Solution {
    public int countValidWords(String sentence) {
        int ans = 0;
        for (String s : sentence.split(" ")) {
            ans += check(s.toCharArray());
        }
        return ans;
    }

    private int check(char[] s) {
        if (s.length == 0) {
            return 0;
        }
        boolean st = false;
        for (int i = 0; i < s.length; ++i) {
            if (Character.isDigit(s[i])) {
                return 0;
            }
            if ((s[i] == '!' || s[i] == '.' || s[i] == ',') && i < s.length - 1) {
                return 0;
            }
            if (s[i] == '-') {
                if (st || i == 0 || i == s.length - 1) {
                    return 0;
                }
                if (!Character.isAlphabetic(s[i - 1]) || !Character.isAlphabetic(s[i + 1])) {
                    return 0;
                }
                st = true;
            }
        }
        return 1;
    }
}

// Accepted solution for LeetCode #2047: Number of Valid Words in a Sentence
func countValidWords(sentence string) (ans int) {
	check := func(s string) int {
		if len(s) == 0 {
			return 0
		}
		st := false
		for i, r := range s {
			if unicode.IsDigit(r) {
				return 0
			}
			if (r == '!' || r == '.' || r == ',') && i < len(s)-1 {
				return 0
			}
			if r == '-' {
				if st || i == 0 || i == len(s)-1 {
					return 0
				}
				if !unicode.IsLetter(rune(s[i-1])) || !unicode.IsLetter(rune(s[i+1])) {
					return 0
				}
				st = true
			}
		}
		return 1
	}
	for _, s := range strings.Fields(sentence) {
		ans += check(s)
	}
	return ans
}

# Accepted solution for LeetCode #2047: Number of Valid Words in a Sentence
class Solution:
    def countValidWords(self, sentence: str) -> int:
        def check(s: str) -> bool:
            st = False
            for i, c in enumerate(s):
                if c.isdigit() or (c in "!.," and i < len(s) - 1):
                    return False
                if c == "-":
                    if (
                        st
                        or i in (0, len(s) - 1)
                        or not s[i - 1].isalpha()
                        or not s[i + 1].isalpha()
                    ):
                        return False
                    st = True
            return True

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

// Accepted solution for LeetCode #2047: Number of Valid Words in a Sentence
struct Solution;

impl Solution {
    fn count_valid_words(sentence: String) -> i32 {
        let mut res = 0;
        for word in sentence.split_whitespace() {
            if Solution::is_valid(word) {
                res += 1;
            }
        }
        res
    }

    fn is_valid(s: &str) -> bool {
        let n = s.len();
        let mut h_count = 0;
        let s: Vec<char> = s.chars().collect();
        for i in 0..n {
            let c = s[i];
            match c {
                'a'..='z' => {}
                ',' | '!' | '.' => {
                    if i != n - 1 {
                        return false;
                    }
                }
                '-' => {
                    h_count += 1;
                    if i == 0 || i == n - 1 {
                        return false;
                    }
                    if !(('a'..='z').contains(&s[i - 1]) && ('a'..='z').contains(&s[i + 1])) {
                        return false;
                    }
                }
                _ => {
                    return false;
                }
            };
        }
        if h_count > 1 {
            return false;
        }
        true
    }
}

#[test]
fn test() {
    let sentence = "cat and  dog".to_string();
    let res = 3;
    assert_eq!(Solution::count_valid_words(sentence), res);
    let sentence = "!this  1-s b8d!".to_string();
    let res = 0;
    assert_eq!(Solution::count_valid_words(sentence), res);
    let sentence = "alice and  bob are playing stone-game10".to_string();
    let res = 5;
    assert_eq!(Solution::count_valid_words(sentence), res);
}

// Accepted solution for LeetCode #2047: Number of Valid Words in a Sentence
function countValidWords(sentence: string): number {
    const check = (s: string): number => {
        if (s.length === 0) {
            return 0;
        }
        let st = false;
        for (let i = 0; i < s.length; ++i) {
            if (/\d/.test(s[i])) {
                return 0;
            }
            if (['!', '.', ','].includes(s[i]) && i < s.length - 1) {
                return 0;
            }
            if (s[i] === '-') {
                if (st || [0, s.length - 1].includes(i)) {
                    return 0;
                }
                if (!/[a-zA-Z]/.test(s[i - 1]) || !/[a-zA-Z]/.test(s[i + 1])) {
                    return 0;
                }
                st = true;
            }
        }
        return 1;
    };
    return sentence.split(/\s+/).reduce((acc, s) => acc + check(s), 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.