Print list items containing all characters of a given word | GeeksforGeeks

There is a list of items. Given a specific word, e.g., “sun”, print out all the items in list which contain all the characters of “sum”

1) Initialize a binary map:
        map[256] = {0, 0, ..}
2) Set values in map[] for the given word "sun"
        map['s'] = 1,  map['u'] = 1,  map['n'] = 1
3) Store length of the word "sun":
        len = 3 for "sun"
4) Pick words (or items)one by one from the list
    a) set count = 0;
    b) For each character ch of the picked word
         if(map['ch'] is set)
            increment count and unset map['ch'] 
    c) If count becomes equal to len (3 for "sun"),
           print the currently picked word.
    d) Set values in map[] for next list item
          map['s'] = 1,  map['u'] = 1,  map['n'] = 1

void print(char *list[], char *word, int list_size)

{

  /*Since calloc is used, map[] is initialized as 0 */

  int *map = (int *)calloc(sizeof(int), NO_OF_CHARS);

  int i, j, count, word_size; 

  /*Set the values in map */

  for (i = 0; *(word+i); i++)

      map[*(word + i)] = 1;

  /* Get the length of given word */

  word_size = strlen(word);

  /* Check each item of list if has all characters

     of word*/

  for (i = 0; i < list_size; i++)

  {

    for(j = 0, count = 0; *(list[i] + j); j++)

    {

      if(map[*(list[i] + j)])

      {

         count++;

          /* unset the bit so that strings like

            sss not printed*/        

         map[*(list[i] + j)] = 0;

      }    

    }

    if(count == word_size)

      printf("\n %s", list[i]);

    /*Set the values in map for next item*/

    for (j = 0; *(word+j); j++)

      map[*(word + j)] = 1;         

  }

}

Read full article from Print list items containing all characters of a given word | GeeksforGeeks