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Word Count

Intro

Given a phrase, count the occurrences of each word in that phrase.

For the purposes of this exercise you can expect that a word will always be one of:

  1. A number composed of one or more ASCII digits (ie "0" or "1234") OR
  2. A simple word composed of one or more ASCII letters (ie "a" or "they") OR
  3. A contraction of two simple words joined by a single apostrophe (ie "it's" or "they're")

When counting words you can assume the following rules:

  1. The count is case insensitive (ie "You", "you", and "YOU" are 3 uses of the same word)
  2. The count is unordered; the tests will ignore how words and counts are ordered
  3. Other than the apostrophe in a contraction all forms of punctuation are ignored
  4. The words can be separated by any form of whitespace (ie "\t", "\n", " ")

For example, for the phrase "That's the password: 'PASSWORD 123'!", cried the Special Agent.\nSo I fled. the count would be:

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that's: 1
the: 2
password: 2
123: 1
cried: 1
special: 1
agent: 1
so: 1
i: 1
fled: 1

Task

Count the occurrences of each word in that phrase.

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#include "word_count.h"
#include "string.h"
#include "stdlib.h"
#include "ctype.h"
#include "stdio.h"

int count_words(const char *sentence, word_count_word_t *words)
{
    int lenght = strlen(sentence);

    char var[MAX_WORD_LENGTH] = {0};

    //strncpy(var, sentence, lenght);
    int i = 0;
    for (i = 0; i < lenght; i++)
    {
        var[i] = tolower(sentence[i]);
    }
    strcat(var, " ");

    for (i = 0; i < (int)strlen(var); i++)
    {
        if (var[i] == '\'' && var[i - 1] == ' ')
        {
            var[i] = ' ';
            continue;
        }
        if (var[i] == ' ' && var[i - 1] == '\'')
        {
            var[i - 1] = ' ';
            continue;
        }
        if ((isalnum(var[i]) == 0 && var[i]!='\'') && isblank(var[i]) == 0)
            var[i] = ' ';
    }

//    puts(var);
    char *var01 = strtok(var, " ");
    int var02 = 0;
    while (var01)
    {
        int ii = 0;
        for (i = 0; i < var02; i++)
        {
            if (strcmp(words[i].text, var01) == 0)
            {
                words[i].count += 1;
                ii = 1;
                break;
            }
        }
        if (ii == 1)
        {
            var01 = strtok(NULL, " ");
            continue;
        }
        strcpy(words[var02].text, var01);
        words[var02].count = 1;
        var01 = strtok(NULL, " ");
        var02++;
    }
    return var02;
}

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#ifndef WORD_COUNT_H
#define WORD_COUNT_H

#define MAX_WORDS 20            // at most MAX_WORDS can be found in the test input string
#define MAX_WORD_LENGTH 50      // no individual word can exceed this length

// results structure
typedef struct word_count_word {
char text[MAX_WORD_LENGTH + 1];      // allow for the string to be null-terminated
int count;
} word_count_word_t;

#define EXCESSIVE_LENGTH_WORD     -1
#define EXCESSIVE_NUMBER_OF_WORDS -2

// count_words - routine to classify the unique words and their frequency in a sentence
// inputs:
//    sentence =  a null-terminated string containing that is analyzed
//
// outputs:
//    words = allocated structure to record the words found and their frequency
//    uniqueWords - number of words in the words structure
//           returns a negative number if an error.
//           words will contain the results up to that point.
int count_words(const char *sentence, word_count_word_t * words);

#endif
Last update: February 8, 2021

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