Program to count the number of words in a sentence

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Write a program to count the number of words in a given sentence. A word is defined as a sequence of characters separated by spaces.

Examples:

Input: "The quick brown fox"
Output: 4
Explanation: The sentence has four words - "The", "quick", "brown", "fox".

Input: "The quick brown fox"
Output: 4
Explanation: The sentence has four words - "The", "quick", "brown", "fox".

Approach: To solve the problem, follow the below idea:

Iterate through each character in the sentence and count the number of spaces to identify the separation between words. The total count of spaces plus 1 gives the number of words.

Step-by-step algorithm:

• Initialize a variable wordCount to 0.
• Iterate through each character in the sentence.
• Check if the current character is a space.
• If a space is found, increment the wordCount.
• After the loop, add 1 to wordCount to account for the last word.
• Output the final value of wordCount.

Below is the implementation of the algorithm:

#include <iostream>

using namespace std;

int countWords(const string& sentence)
{
    // Initialize count to 1 assuming at least one word is
    // present
    int count = 1;

    for (char character : sentence) {
        if (character == ' ') {
            count++;
        }
    }

    return count;
}

int main()
{
    string sentence = "Programming is fun and challenging";
    cout << "Number of words: " << countWords(sentence)
         << endl;

    return 0;
}

#include <stdio.h>

int countWords(char sentence[]) {
    int count = 1;  // Initialize count to 1 assuming at least one word is present

    for (int i = 0; sentence[i] != '\0'; ++i) {
        if (sentence[i] == ' ') {
            count++;
        }
    }

    return count;
}

int main() {
    char sentence[] = "Programming is fun and challenging";
    printf("Number of words: %d\n", countWords(sentence));

    return 0;
}

public class WordCounter {
    static int countWords(String sentence) {
        // Initialize count to 1 assuming at least one word is present
        int count = 1;

        for (int i = 0; i < sentence.length(); ++i) {
            if (sentence.charAt(i) == ' ') {
                count++;
            }
        }

        return count;
    }

    public static void main(String[] args) {
        String sentence = "Programming is fun and challenging";
        System.out.println("Number of words: " + countWords(sentence));
    }
}

def count_words(sentence):
    # Initialize count to 1 assuming at least one word is present
    count = 1

    for char in sentence:
        if char == ' ':
            count += 1

    return count

# Example usage
sentence = "Programming is fun and challenging"
print("Number of words:", count_words(sentence))

using System;

class Program {
    static int CountWords(string sentence)
    {
        // Initialize count to 1 assuming at least one word
        // is present
        int count = 1;

        foreach(char character in sentence)
        {
            if (character == ' ') {
                count++;
            }
        }

        return count;
    }

    static void Main()
    {
        string sentence
            = "Programming is fun and challenging";
        Console.WriteLine("Number of words: "
                          + CountWords(sentence));
    }
}

function countWords(sentence) {
    // Initialize count to 1 assuming at least one word is present
    let count = 1;

    for (let char of sentence) {
        if (char === ' ') {
            count++;
        }
    }

    return count;
}

// Example usage
let sentence = "Programming is fun and challenging";
console.log("Number of words:", countWords(sentence));

Number of words: 5

Time Complexity: O(N) where N is the length of the sentence.
Auxiliary Space: O(1)