Word Counter - Problem

Count the number of words in a sentence. A word is defined as a sequence of non-space characters.

Important: Handle multiple consecutive spaces between words and leading/trailing spaces properly.

For example:

"hello world" has 2 words
" hello world " has 2 words (ignore extra spaces)
"" has 0 words
" " has 0 words

Input & Output

Example 1 — Basic Case

$ Input: sentence = "hello world"

› Output: 2

💡 Note: Two words separated by a single space: "hello" and "world"

Example 2 — Multiple Spaces

$ Input: sentence = " hello world "

› Output: 2

💡 Note: Same two words but with leading, trailing, and multiple spaces between - still counts as 2 words

Example 3 — Empty String

$ Input: sentence = ""

› Output: 0

💡 Note: Empty string contains no words

Constraints

0 ≤ sentence.length ≤ 10⁴
sentence consists of English letters, digits, and spaces ' '

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 15

The key insight is to handle multiple spaces properly by either tracking word boundaries manually or using built-in split methods. Best approach is the split method for simplicity and reliability. Time: O(n), Space: O(k) where k is number of words.

Common Approaches

✓ Built-in Split Method

⏱️ Time: O(n) Space: O(k)

Leverage the built-in split() method which automatically handles multiple spaces and returns only non-empty segments. Simply count the resulting array length.

Character by Character Scan

⏱️ Time: O(n) Space: O(1)

Scan every character in the sentence one by one. Keep a flag to track whether we're currently inside a word or in a space. When transitioning from space to non-space, increment word count.

Built-in Split Method — Algorithm Steps

Call split() method on the sentence with space as delimiter
The method automatically handles multiple consecutive spaces
Filter out any empty strings (some languages need this)
Return the length of the resulting word array

Visualization

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Step-by-Step Walkthrough

Input String

Original sentence with various spacing

Split Operation

Built-in method separates by whitespace

Count Words

Return length of resulting array

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <ctype.h>

int solution(char* sentence) {
    if (sentence == NULL) {
        return 0;
    }
    
    int count = 0;
    int i = 0;
    int len = strlen(sentence);
    
    while (i < len) {
        // Skip spaces
        while (i < len && isspace(sentence[i])) {
            i++;
        }
        
        // If we found a word
        if (i < len) {
            count++;
            // Skip the word
            while (i < len && !isspace(sentence[i])) {
                i++;
            }
        }
    }
    
    return count;
}

int main() {
    char sentence[1000];
    fgets(sentence, sizeof(sentence), stdin);
    sentence[strcspn(sentence, "\n")] = 0;
    int result = solution(sentence);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Split method needs to scan through all n characters once

✓ Linear Growth

Space Complexity

O(k)

Creates array to store k words, where k is number of words found

✓ Linear Space

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Word Counter - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Built-in Split Method — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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