C++ function to count all the words in a string

Question

I was asked this during an interview and apparently it's an easy question but it wasn't and still isn't obvious to me.

Given a string, count all the words in it. Doesn't matter if they are repeated. Just the total count like in a text files word count. Words are anything separated by a space and punctuation doesn't matter, as long as it's part of a word.

For example: A very, very, very, very, very big dog ate my homework!!!! ==> 11 words

My "algorithm" just goes through looking for spaces and incrementing a counter until I hit a null. Since i didn't get the job and was asked to leave after that I guess My solution wasn't good? Anyone have a more clever solution? Am I missing something?

Answer 1

Assuming words are white space separated:

unsigned int countWordsInString(std::string const& str)
{
    std::stringstream stream(str);
    return std::distance(std::istream_iterator<std::string>(stream), std::istream_iterator<std::string>());
}

Note: There may be more than one space between words. Also this does not catch other white space characters like tab new line or carriage return. So counting spaces is not enough.

The stream input operator >> when used to read a string from a stream. Reads one white space separated word. So they were probably looking for you to use this to identify words.

std::stringstream  stream(str);
std::string        oneWord;

stream >> oneWord; // Reads one space separated word.

When can use this to count words in a string.

std::stringstream  stream(str);
std::string        oneWord;
unsigned int       count = 0;

while(stream >> oneWord) { ++count;}
// count now has the number of words in the string.

Getting complicated:
Streams can be treated just like any other container and there are iterators to loop through them std::istream_iterator. When you use the ++ operator on an istream_iterator it just read the next value from the stream using the operator >>. In this case we are reading std::string so it reads a space separated word.

std::stringstream  stream(str);
std::string        oneWord;
unsigned int       count = 0;

std::istream_iterator loop = std::istream_iterator<std::string>(stream);
std::istream_iterator end  = std::istream_iterator<std::string>();

for(;loop != end; ++count, ++loop) { *loop; }

Using std::distance just wraps all the above in a tidy package as it find the distance between two iterators by doing ++ on the first until we reach the second.

To avoid copying the string we can be sneaky:

unsigned int countWordsInString(std::string const& str)
{
    std::stringstream stream;

    // sneaky way to use the string as the buffer to avoid copy.
    stream.rdbuf()->pubsetbuf (str.c_str(), str.length() );
    return std::distance(std::istream_iterator<std::string>(stream), std::istream_iterator<std::string>());
}

Note: we still copy each word out of the original into a temporary. But the cost of that is minimal.

Answer 2

A less clever, more obvious-to-all-of-the-programmers-on-your-team method of doing it.

#include <cctype>

int CountWords(const char* str)
{
   if (str == NULL)
      return error_condition;  // let the requirements define this...

   bool inSpaces = true;
   int numWords = 0;

   while (*str != NULL)
   {
      if (std::isspace(*str))
      {
         inSpaces = true;
      }
      else if (inSpaces)
      {
         numWords++;
         inSpaces = false;
      }

      ++str;
   }

   return numWords;
}

Answer 3

This can be done without manually looking at every character or copying the string.

#include <boost/iterator/transform_iterator.hpp>
#include <cctype>

boost::transform_iterator
    < int (*)(int), std::string::const_iterator, bool const& >
    pen( str.begin(), std::isalnum ), end( str.end(), std::isalnum );

size_t word_cnt = 0;

while ( pen != end ) {
    word_cnt += * pen;
    pen = std::mismatch( pen+1, end, pen ).first;
}

return word_cnt;

I took the liberty of using isalnum instead of isspace.

This is not something I would do at a job interview. (It's not like it compiled the first time.)

Or, for all the Boost haters ;v)

if ( str.empty() ) return 0;

size_t word_cnt = std::isalnum( * str.begin() );

for ( std::string::const_iterator pen = str.begin(); ++ pen != str.end(); ) {
    word_cnt += std::isalnum( pen[ 0 ] ) && ! std::isalnum( pen[ -1 ] );
}

return word_cnt;

Answer 4

Another boost based solution that may work (untested):

vector<string> result;
split(result, "aaaa bbbb cccc", is_any_of(" \t\n\v\f\r"), token_compress_on);

More information can be found in the Boost String Algorithms Library