Data Modelling & AI Data Structure & Algorithm

Check if the given string is valid English word or not

By Umr Jansen

23 June 2025

0

Given string str, the task is to check if this string str consists of valid English words or not.

A string is known as a valid English word if it meets all the below criteria-

The string can have an uppercase character as the first character only.
The string can only have lowercase characters.
The string can consist of only one hyphen(‘-‘) surrounded by characters on both ends.
The string cannot consist of any digits.
If there is any punctuation mark it must be only one and it must be present at the end.

Print the number of valid words in the string str.

Input: str = “i Love- Geeks-forneveropen!”
Output: 1 word
Explanation:
word 1 = “i” does not contain first uppercase character, it is not valid word
word 2 = “Love-” hyphen is not surrounded by characters on both ends, it is not valid word
word 3 = “Geeks-forneveropen!” is a valid word

Input: str = “!this 1-s b8d!”
Output: 0 words
Explanation:
word 1 = “!this” punctuation mark is in the beginning, it is not valid word
word 2 = “1-s” digit as first character, it is not valid word
word 3 = “b8d!” first character is not uppercase, it is not valid word

Approach:

Initialize the variable ans to keep count of the number of valid words.
Loop through each word present in the sentence.
Check each letter of the word to see if it meets the criteria mentioned in the problem statement.
If any of the criteria is not met then return false.
If all the criteria are satisfied by the word, then increment the value of the variable ans.
Print the value of the variable ans.

Below is the C++ program of the above approach-

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find valid words
bool ValidWords(string sentence)
{
    int hyphen = 0;
    int size = sentence.size();
    if (isupper(sentence[0])) {
        for (int i = 0; i < size; i++) {
            // Check for numbers
            if (isdigit(sentence[i]))
                return false;
            if (isupper(sentence[i]))
                return false;
 
            if (isalpha(sentence[i]))
                continue;
 
            if (sentence[i] == '-') {
                // Only 1 hyphen is allowed
                if (++hyphen > 1)
                    return false;
 
                // hyphen should be surrounded
                // by letters
                if (i - 1 < 0
                    || !isalpha(sentence[i - 1])
                    || i + 1 >= size
                    || !isalpha(sentence[i + 1]))
                    return false;
            }
 
            // Punctuation must be at the
            // end of the word
            else if (i != size - 1
                     && ispunct(sentence[i]))
                return false;
        }
    }
    else
        return true;
}
 
// Driver code
int main()
{
    string sentence = "i Love- Geeks-Forneveropen!";
 
    istringstream s(sentence);
    string word;
    int ans = 0;
 
    while (s >> word)
        if (ValidWords(word))
            ans++;
 
    // Display the result
    cout << ans << " words";
}

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
 
    // Function to find valid words
    static boolean ValidWords(String sentence)
    {
        int hyphen = 0;
        int size = sentence.length();
        if (Character.isUpperCase(sentence.charAt(0))) {
            for (int i = 0; i < size; i++) {
                // Check for numbers
                if (Character.isDigit(sentence.charAt(i)))
                    return false;
                if (Character.isUpperCase(
                        sentence.charAt(i)))
                    return false;
 
                if (Character.isAlphabetic(
                        sentence.charAt(i)))
                    continue;
 
                if (sentence.charAt(i) == '-') {
                    // Only 1 hyphen is allowed
                   hyphen = hyphen +1 ;
                    if (hyphen > 1)
                        return false;
 
                    // hyphen should be surrounded
                    // by letters
                    if (i - 1 < 0
                        || !Character.isAlphabetic(
                            sentence.charAt(i - 1))
                        || i + 1 >= size
                        || !Character.isAlphabetic(
                            sentence.charAt(i + 1)))
                        return false;
                }
 
                // Punctuation must be at the
                // end of the word
                else if (i != size - 1
                         && ((sentence.charAt(i) == '!'
                              || sentence.charAt(i) == ','
                              || sentence.charAt(i) == ';'
                              || sentence.charAt(i) == '.'
                              || sentence.charAt(i) == '?'
                              || sentence.charAt(i) == '-'
                              || sentence.charAt(i) == '\''
                              || sentence.charAt(i) == '\"'
                              || sentence.charAt(i)
                                     == ':')))
                    return false;
            }
        }
        else
            return true;
      return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String sentence = "i Love- Geeks-Forneveropen!";
        int ans = 0;
        String words[] = sentence.split(" ");
        for (String word : words) {
            if (ValidWords(word)==true){
                ans++;
            }
        }
 
        // Display the result
        System.out.print(ans + " words");
    }
}

Python3

# Python program to implement
# the above approach
 
# Function to find valid words
def ValidWords(sentence):
    hyphen = 0
    size = len(sentence)
    if (sentence[0] >= 'A' and sentence[0] <= 'Z'):
        for i in range(size):
           
            # Check for numbers
            if (sentence[i] >= '0' and sentence[i] <= '9'):
                return False
            if (sentence[i] >= 'A' and sentence[i] <= 'Z'):
                return False
            if (sentence[i] >= 'a' and sentence[i] <= 'z' or sentence[i] >= 'A'
                and sentence[i] <= 'Z'):
                continue
            if (sentence[i] == '-'):
               
                # Only 1 hyphen is allowed
                if (hyphen+1 > 1):
                    return False
                   
                # hyphen should be surrounded
                # by letters
                if (i - 1 < 0 or ~(sentence[i - 1] >= 'a' and
                    sentence[i - 1] <= 'z' or sentence[i - 1] >= 'A'
                    and sentence[i - 1] <= 'Z') or i + 1 >= size or
                    ~(sentence[i + 1] >= 'a' and sentence[i + 1] <= 'z'
                      or sentence[i + 1] >= 'A' and sentence[i + 1] <= 'Z')):
                    return False
 
            # Punctuation must be at the
            # end of the word
            elif (i != size - 1 and ((sentence[i] == '!' or sentence[i] == ','
                   or sentence[i] == ';' or sentence[i] == '.' or sentence[i] == '?'
                   or sentence[i] == '-' or sentence[i] == '\''or sentence[i] == '\"'
                   or sentence[i] == ':'))):
                return False
    else:
        return True
 
# Driver code
sentence = "i Love- Geeks-Forneveropen!"
 
word = sentence.split(' ')
ans = 0
 
for indx in word :
    if (ValidWords(indx)):
        ans += 1
 
# Display the result
print(f"{ans} words")
 
# This code is contributed by shinjanpatra

C#

/*package whatever //do not write package name here */
using System;
 
class GFG
{
 
  // Function to find valid words
  static bool ValidWords(String sentence)
  {
    int hyphen = 0;
    int size = sentence.Length;
    if (char.IsUpper(sentence[0]))
    {
      for (int i = 0; i < size; i++)
      {
         
        // Check for numbers
        if (char.IsDigit(sentence[i]))
          return false;
        if (char.IsUpper(sentence[i]))
          return false;
 
        if (char.IsLetter(sentence[i]))
          continue;
 
        if (sentence[i] == '-')
        {
           
          // Only 1 hyphen is allowed
          hyphen = hyphen + 1;
          if (hyphen > 1)
            return false;
 
          // hyphen should be surrounded
          // by letters
          if (i - 1 < 0
              || !char.IsLetter(sentence[i - 1])
              || i + 1 >= size
              || !char.IsLetter(sentence[i + 1]))
            return false;
        }
 
        // Punctuation must be at the
        // end of the word
        else if (i != size - 1
                 && ((sentence[i] == '!'
                      || sentence[i] == ','
                      || sentence[i] == ';'
                      || sentence[i] == '.'
                      || sentence[i] == '?'
                      || sentence[i] == '-'
                      || sentence[i] == '\''
                      || sentence[i] == '\"'
                      || sentence[i]
                      == ':')))
          return false;
      }
    }
    else
      return true;
    return false;
  }
 
  // Driver code
  public static void Main()
  {
    String sentence = "i Love- Geeks-Forneveropen!";
    int ans = 0;
    String[] words = sentence.Split(" ");
    foreach (String word in words)
    {
      if (ValidWords(word) == true)
      {
        ans++;
      }
    }
 
    // Display the result
    Console.Write(ans + " words");
  }
}
 
// This code is contributed by gfgking.

Javascript

<script>
    // JavaScript program to implement
    // the above approach
 
    // Function to find valid words
    const ValidWords = (sentence) => {
        let hyphen = 0;
        let size = sentence.length;
        if (sentence[0] >= 'A' && sentence[0] <= 'Z')
        {
            for (let i = 0; i < size; i++) 
            {
                // Check for numbers
                if (sentence[i] >= '0' && sentence[i] <= '9')
                    return false;
                if (sentence[i] >= 'A' && sentence[i] <= 'Z')
                    return false;
 
                if (sentence[i] >= 'a' && sentence[i] <= 'z' || 
                sentence[i] >= 'A' && sentence[i] <= 'Z')
                    continue;
 
                if (sentence[i] == '-') {
                    // Only 1 hyphen is allowed
                    if (++hyphen > 1)
                        return false;
 
                    // hyphen should be surrounded
                    // by letters
                    if (i - 1 < 0
                        || !(sentence[i - 1] >= 'a' && 
                        sentence[i - 1] <= 'z' || 
                        sentence[i - 1] >= 'A' && 
                        sentence[i - 1] <= 'Z')
                        || i + 1 >= size
                        || !(sentence[i + 1] >= 'a' && 
                        sentence[i + 1] <= 'z' || 
                        sentence[i + 1] >= 'A' && 
                        sentence[i + 1] <= 'Z'))
                        return false;
                }
 
                // Punctuation must be at the
                // end of the word
                else if (i != size - 1
                    && ((sentence[i] == '!'
                        || sentence[i] == ','
                        || sentence[i] == ';'
                        || sentence[i] == '.'
                        || sentence[i] == '?'
                        || sentence[i] == '-'
                        || sentence[i] == '\''
                        || sentence[i] == '\"'
                        || sentence[i]
                        == ':')))
                    return false;
            }
        }
        else
            return true;
    }
 
    // Driver code
 
    let sentence = "i Love- Geeks-Forneveropen!";
 
    let word = sentence.split(' ');
    let ans = 0;
 
    for (let indx in word)
        if (ValidWords(word[indx]))
            ans++;
 
    // Display the result
    document.write(`${ans} words`);
 
    // This code is contributed by rakeshsahni
 
</script>

Output

1 words

Time Complexity: O(N) as only one traversal of the string of length N is enough for the algorithm to perform all the tasks hence the overall complexity is linear.
Auxiliary Space: O(N) as the variable s stores all the words of the strings hence the overall space occupied by the algorithm is equal to the length of the string

Approach 2:

Here is another approach to check for valid words in a given sentence:

Split the sentence into words using whitespace as a delimiter.
For each word, iterate over its characters and perform the following checks:
The first character should be an uppercase or lowercase letter.
All other characters should be lowercase letters or hyphens.
There should be at most one hyphen, and it should not be the first or last character.
There should be no digits or punctuation marks.
If a word passes all the checks, increment a counter to keep track of the number of valid words.

After processing all the words in the sentence, display the number of valid words found.
Here is an implementation of this approach in C++:

C++

#include <iostream>
#include <string>
#include <cctype>
 
using namespace std;
 
bool isValidWord(const string& word) {
    if (word.empty()) {
        return false;
    }
 
    if (!isalpha(word[0])) {
        return false;
    }
 
    int hyphenCount = 0;
 
    for (int i = 1; i < word.length(); i++) {
        if (word[i] == '-') {
            hyphenCount++;
 
            if (hyphenCount > 1 || i == 1 || i == word.length() - 1 || !isalpha(word[i - 1]) || !isalpha(word[i + 1])) {
                return false;
            }
        }
        else if (!islower(word[i])) {
            return false;
        }
    }
 
    return true;
}
 
int countValidWords(const string& sentence) {
    int count = 0;
    string word;
 
    for (int i = 0; i < sentence.length(); i++) {
        if (isspace(sentence[i])) {
            if (isValidWord(word)) {
                count++;
            }
 
            word.clear();
        }
        else {
            word += sentence[i];
        }
    }
 
    if (isValidWord(word)) {
        count++;
    }
 
    return count;
}
 
int main() {
    string sentence = "i Love- Geeks-Forneveropen!";
 
    int validWordCount = countValidWords(sentence);
 
    cout << validWordCount << " words" << endl;
 
    return 0;
}

Java

import java.util.*;
 
public class ValidWordsCount {
 
    // Function to check if a given word is valid
    public static boolean isValidWord(String word) {
        if (word.isEmpty()) {
            return false;
        }
 
        // Check if the first character is a letter
        if (!Character.isLetter(word.charAt(0))) {
            return false;
        }
 
        int hyphenCount = 0;
 
        // Loop through the characters of the word starting from the second character
        for (int i = 1; i < word.length(); i++) {
            if (word.charAt(i) == '-') {
                hyphenCount++;
 
                // Conditions to check if hyphen placement is valid
                if (hyphenCount > 1 || i == 1 || i == word.length() - 1 || !Character.isLetter(word.charAt(i - 1)) || !Character.isLetter(word.charAt(i + 1))) {
                    // If there is more than one hyphen, if it's at the beginning or end of the word, or if the characters around the hyphen are not letters, the word is invalid.
                    return false;
                }
            } else if (!Character.isLowerCase(word.charAt(i))) {
                // If a non-lowercase letter is found (other than a hyphen), the word is invalid.
                return false;
            }
        }
 
        // If all conditions pass, the word is valid.
        return true;
    }
 
    // Function to count the number of valid words in a sentence
    public static int countValidWords(String sentence) {
        int count = 0;
        String word = "";
 
        // Loop through the characters of the sentence
        for (int i = 0; i < sentence.length(); i++) {
            if (Character.isWhitespace(sentence.charAt(i))) {
                // If a whitespace character is encountered, check if the current word is valid and increment the count if it is.
                if (isValidWord(word)) {
                    count++;
                }
 
                // Reset the word variable for the next word
                word = "";
            } else {
                // If a non-whitespace character is encountered, append it to the current word.
                word += sentence.charAt(i);
            }
        }
 
        // After the loop, check the last word in the sentence and increment the count if it is valid.
        if (isValidWord(word)) {
            count++;
        }
 
        // Return the total count of valid words in the sentence.
        return count;
    }
 
    public static void main(String[] args) {
        // Input sentence
        String sentence = "i Love- Geeks-Forneveropen!";
 
        // Call the function to count the number of valid words in the sentence
        int validWordCount = countValidWords(sentence);
 
        // Print the result
        System.out.println(validWordCount + " words");
    }
}

Output

1 words

Time Complexity: O(N) where N is the of length of string.
Auxiliary Space: O(1) for constant time taken

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Check if the given string is valid English word or not

C++

Java

Python3

C#

Javascript

C++

Java

Is MCP Already Outdated? The Real Reason Anthropic Shipped Skills—and How to Pair Them with Milvus

Unlocking 8× Milvus Performance with Cloudian HyperStore and NVIDIA RDMA for S3 Storage

Power high performance RAG for GenAI with HPE Alletra Storage MP + Milvus

LEAVE A REPLY Cancel reply

Most Popular

This Garmin powerhouse outclasses every ‘Ultra’ watch, and calling it a must-buy feels like an understatement

Wear OS 6 lands on the Samsung’s Galaxy Watch 6 and Watch 6 Classic

T-Mobile will now start charging for a perk that used to be free

Spotify now lets you transfer playlists from other services

EDITOR PICKS

This Garmin powerhouse outclasses every ‘Ultra’ watch, and calling it a must-buy feels like an understatement

Wear OS 6 lands on the Samsung’s Galaxy Watch 6 and Watch 6 Classic

T-Mobile will now start charging for a perk that used to be free

POPULAR POSTS

This Garmin powerhouse outclasses every ‘Ultra’ watch, and calling it a must-buy feels like an understatement

Wear OS 6 lands on the Samsung’s Galaxy Watch 6 and Watch 6 Classic

T-Mobile will now start charging for a perk that used to be free

POPULAR CATEGORY

ABOUT US

FOLLOW US