CODEWITHSUNDEEP
carrayscharcompare
user1009835
user1009835

Reputation:

Compare char arrays

If I had an array of characters for example:

A = [w, o, r, n, g, , w, o, r, d]

And another array for example:

B = [c, o, r, r, e, c, t, , w, o, r, d, .]

I need to compare the words in array A (which are separated by a blank space) to array B and if any of the words in the first array exist in the second array, then that word should be printed. So for example, since "word" exists in the first array and in the second array, then "word" should be printed out.

How do I go about this?

Upvotes: 2

Views: 3550

Answers (3)

xanatos
xanatos

Reputation: 111860

Let's see how I would do it:

You will need a function that, given an array of char, splits it in an array of words (and put them in C strings, NUL terminated please :-) ). I would put the length of this array and the array in a struct

struct WordCollection
{
    size_t NumWords;
    char **Words;
}

Now... How to do this function?

Let's say we "cheat" a little and decide that our arrays A and B are NUL terminated (or if they are . terminated like B, then you replace the . with a NUL). Now, this being C, you should first count the number of spaces in the string, allocate an array of char* (WordCollection::Words) big enough to contain n + 1 char* (and put this n + 1 in WordCollection::NumWords) and using strtok "tokenize" the string and put the words in the array you created.

Then you should (could) split the A and B array in words using this function. You'll obtain two WordCollection, A1 and B1.

To make it quicker, I would qsort B1.

Then for each word in A1 you bsearch it in B1 (it isn't a bad word... It means Binary Search, and it's a quick method of searching something in an ordered array)

Done :-)

I'll add that, if this is the first time you use bsearch and qsort, it's better you look at the samples you can find around. Their syntax can be "tricky".

Now... I know you won't look at the code :-) so I'll put it here

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

struct WordCollection
{
    size_t NumWords;
    char **Words;
};

void splitWord(char *str, struct WordCollection *wc)
{
    char *c;
    char **currentWord;

    c = str;

    wc->NumWords = 1;

    while (*c != '.')
    {
        if (*c == ' ')
        {
            wc->NumWords++;
        }

        c++;
    }

    *c = '\0';

    wc->Words = (char**)malloc(wc->NumWords * sizeof(char*));

    c = strtok(str, " ");

    currentWord = wc->Words;

    while (c)
    {
        *currentWord = c;
        currentWord++;

        c = strtok(NULL, " ");
    }
}

int myComp(const void *p1, const void *p2)
{
    return strcmp(*(const char**)p1, *(const char**)p2);
}

int main(void)
{
    char a[] = { 'w', 'o', 'r', 'n', 'g', ' ', 'w', 'o', 'r', 'd', '.' };
    char b[] = { 'c', 'o', 'r', 'r', 'e', 'c', 't', ' ', 'w', 'o', 'r', 'd', '.' };

    struct WordCollection a1, b1;
    struct WordCollection *pSmaller, *pBigger;

    size_t i;

    splitWord(a, &a1);
    splitWord(b, &b1);

    if (a1.NumWords <= b1.NumWords)
    {
        pSmaller = &a1;
        pBigger = &b1;
    }
    else
    {
        pSmaller = &b1;
        pBigger = &a1;
    }

    qsort(pBigger->Words, pBigger->NumWords, sizeof(char*), myComp);

    for (i = 0; i < pSmaller->NumWords; i++)
    {
        void *res = bsearch(&pSmaller->Words[i], pBigger->Words, pBigger->NumWords, sizeof(char*), myComp);
        if (res)
        {
            printf("Found: %s", pSmaller->Words[i]);
        }
    }

    free(a1.Words);
    free(b1.Words);

    return 0;
}

And on ideone

Upvotes: 3

Agnius Vasiliauskas
Agnius Vasiliauskas

Reputation: 11267

You can also do it like that:

  1. Insert all words from set A into set C with suffix 'A'. You will get =>
    worngA
    wordA

  2. Insert all words from set B into set C with suffix 'B'. You will get =>
    correctB
    wordB

  3. Run sorting algo on set C such as qsort. You will get =>
    correctB
    wordA
    wordB
    worngA

  4. Loop in set C until it's size-1. Compare word[i] with word[i+1] - if they match except the last letter - you found duplicate and you can print it out.

I don't know about this algorithm complexity, but it clearly should be faster than just brute-force scan of all words combinations :-)

Upvotes: 0

Hot Licks
Hot Licks

Reputation: 47729

Basically you need to somehow separate the words, and then iterate through the combinations. There are a hundred ways to do this -- it simply requires programming.

Upvotes: 1

Related Questions