Find Subsets of Given String

Method I:

       Can find substring for any length of string because we are taking binary number as string while Conversion itself

#include<stdio.h>

#include<string.h>

#include<math.h>

char *GetBinary(int dec);

int main()

{

 int i,j,len,count=0,no_of_zeros;

 char str[20],temp[20],binary[20];

 printf("Enter a string\n");

 gets(str);

 len=strlen(str);

 printf("Subsets of Given String are");

 for(i=0;i<(pow(2,len));i++)//2^len is the number of subsets for given length.

 {

  strcpy(binary,GetBinary(i));

  no_of_zeros=len-strlen(binary);//Since zeros will not be there before single digits and double etc depending on length of binary number.

 //To compensate them we used here

  for(j=0;j<len;j++)

  {

         if(j>=noofzeros)

        {

          if(binary[j-no_of_zeros]!='0')

           printf("%c",str[j]);

        }

  }

        printf("\n");

 }

 return(0);

}

char *GetBinary(int dec)

{

 int temp,i,j=1,k=0;

 char binary[20];

 temp=dec;

 while(temp!=0)

 {

   i=temp%2;

      binary[k]=i+'0';

   temp=temp/2;

      j=j*10;

      k++;

 }

 binary[k]='\0';

 return(strrev(binary));

}

Method II:

       Can find subsets for smaller strings as we are taking binary number as integer

 

#include<stdio.h>

#include<string.h>

#include<math.h>

int GetBinary(int dec);

int main()

{

 int i,j,binary,len,count=0,tempLen;

 char str[20],temp[20];

 printf("Enter a string\n");

 gets(str);

 len=strlen(str);

 for(i=1;i<(pow(2,len));i++)//2^len is the number of subsets for given length

 {

  binary=GetBinary(i);

  sprintf(temp, "%d", binary);//To convert integer to string

  tempLen=len-strlen(temp);

  for(j=0;j<len;j++)

  {

   if(j>=tempLen)

   {

    if(temp[j-tempLen]!='0')

       printf("%c",str[j]);

   }

  }

  printf("\n");

 }
return 0;

}

int GetBinary(int dec)

{

 int temp,i,j=1,binary=0;

 temp=dec;

 while(temp!=0)

 {

   i=temp%2;

      binary=binary+(i*j);

   temp=temp/2;

      j=j*10;

 }

 return(binary);

}

Input :

Tea

Output:

T
e 
a
Te
Ta
ea
Tea

Input :
Help
Output :
H
e
l
p
He
Hl
Hp
el
ep
lp
Hel
Hep
Hlp
elp
Help