Vfrdtyky

How can I complete the function canArrangeWords() ?

Question : Given a set of words check if we can arrange them in a list such that the last letter of any word and first letter of another word are same. The input function canArrangeWords shall contain an integer num and array of words arr. num denotes the number of word in the list (1<=num<=100). arr shall contain words consisting of lower case letters between 'a' - 'z' only . return 1 if words can be arranged in that fashion and -1 if cannot.

Input : 4  pot ten nice eye

output : 1

input : 3 fox owl pond 

output: -1

Please help me complete this program .
**

#include<stdio.h>

#include<string.h>

int canArrangewords(int,char [100][100]);





void main(){

    int n ,count=0 , i ; 

    char arrayS[100][100];

    scanf("%d",&n);

    for (i = 0; i < n; ++i)

    {

        scanf("%s",arrayS[i]);



    }



    for(i=0;i<n;i++)

    {

        printf("%s",arrayS[i]);

        printf("n");



    }

    printf("%cn",arrayS[2][4]);



    canArrangewords(n , arrayS);



}



int canArrangewords(int n,char arrayS[100][100]){





    int i , j ;

    for ( i = 0; i < n; i++)

    {

        for ( j = i+1 ; j < strlen(arrayS[j+1]); i++)

        {

            int flag = strlen(arrayS[j+1]) - 1;

            int temp = strcmp(arrayS[i][0],arrayS[j][flag]);







            }



        }





    }







}

Well, first of all think of the way you can reach that answer.

If you only need to know if they can or can not be arranged and you do not have to do so your self you can use an empty array of int array[26] for each letter a-z.

The rule is that from all the first and last letters for all the words only two MAY appear an odd amount of times - the first letter of first word in list and the last letter in the last word in the list, the rest MUST appear an even amount of times. I would add a check to make sure the letters are lowercase as well. good luck!

#include <stdio.h>

#include <stdlib.h>

#include <string.h>



#define MINASCII 97

#define LETTERS 26

void UpdateArray(char letter, int* arr)

{



  if(arr[letter - MINASCII] == 0)

  {

      ++arr[letter - MINASCII];

  }

  else

  {

     --arr[letter - MINASCII];/*for each second time same letter is seen reduce back to zero */

  }

}

int canArrangewords(int wordNum, char* wordArr)

{

  int arr[LETTERS] = {0};

  int i = 0;

  int count = 0 ;

  char first;

  char last;

  char* string;

  for (i= 0; i< wordNum; ++i)

  {

     string = wordArr[i];

     first = string[0];

     last = string[strlen(string)-1];

     UpdateArray(first, &arr[0]);

     UpdateArray(last, &arr[0]);

 }



  for(i = 0; i< LETTERS; ++i)

  {

    count+=arr[i];

  }

  if(count == 2 || count == 0)/*either once each or twice -see word1 example in main*/

  {

    return 1;

  }

  return -1;

}





int main()

{

  int i = 0;

  char* words = {"pot", "ten", "nice", "eye"};

  char* words1 = {"pot", "ten", "nip"};

  char* words2 = {"fox", "owl", "pond"};

  i = canArrangewords(4,words);

  printf("%dn", i);

  i = canArrangewords(3,words1);

  printf("%dn", i);

  i = canArrangewords(3,words2);

  printf("%dn", i);

  return 0;

 }

Change your array of words into an array of pointers to words. Then you can simply exchange the pointers.

To speed things up, instead of a pointer to a word, have it point to a structure:

struct WORD {

    char *firstchar;   // begin of word

    char *lastchar;    // last char of word

} *words[100];         // array of 100 pointers to words

To read the words:

char buf[100];

for (i = 0; i < n; ++i)

{

    scanf("%s",buf);

    int len= strlen(buf);

    words[i]= malloc(sizeof(struct WORDS));

    words[i]->firstchar= malloc(len+1);

    strcpy(words[i]->firstchar, buf);

    words[i]->lastchar= words[i]->firstchar + len-1;

}

Now compare and sort:

        if (*words[i]->lastchar == *words[j]->firstchar) {

            struct WORDS *tmp= words[i+1];

            words[i+1]= words[j];

            words[j]= tmp;

        }

Do this in a loop, a kind of bubble sort. I leave that to you.