Word Break Problem

Reference:

1. GeeksforGeeks: Word Break Problem

Recursive version with c(20170101):

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

#define MAX_STRING 256
#define IS_EQUAL 0

int dictionaryContains(char *str)
{
    char *dictionary[] = { "mobile","samsung","sam","sung","man","mango",
        "icecream","and","go","i","like","ice","cream" };

    int size = sizeof(dictionary) / sizeof(dictionary[0]);

    for (int i = 0; i < size; i++)
        if (IS_EQUAL == strcmp(dictionary[i], str))
            return true;

    return false;
}

bool wordBreak(char *str)
{
    int nIndex = 0;
    int nStrLen = 0;

    nStrLen = strlen(str);

    if (nStrLen == 0)
        return true;

    for (nIndex =1; nIndex<=nStrLen; nIndex++)
    {
        char sTargetStr[MAX_STRING] = { 0 };
        char sSecondStr[MAX_STRING] = { 0 };
        strncpy(sTargetStr, str, nIndex);
        strncpy(sSecondStr, str+nIndex, nStrLen-nIndex);

        if (dictionaryContains(sTargetStr) && wordBreak(sSecondStr))
            return true;
    }
    return false;
}

int main()
{
    wordBreak("ilike") ? printf("Yes\n") : printf("No\n");                    //yes
    wordBreak("iiiiiii") ? printf("Yes\n") : printf("No\n");                  //yes
    wordBreak("") ? printf("Yes\n") : printf("No\n");                         //yes
    wordBreak("ilikelikeimangoiii") ? printf("Yes\n") : printf("No\n");       //yes
    wordBreak("samsungandmango") ? printf("Yes\n") : printf("No\n");          //yes
    wordBreak("samsungandmangok") ? printf("Yes\n") : printf("No\n");         //no

    return 0;
}

Iterative version with c:

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