Hi guys need one help from you. Please solve this question?

Question

You are given a string, without any duplicate characters. For example: “abcd”. Write a function that takes such a string as input and prints all possible permutations of the string

Note: Write programs for the above questions in any programming language of your choice

Answer 1

the following is in C++



/*guys..I have implemented Jhonson trotter algorithm..u can

print permutations of 123..n. implement the same for strings!*/





#include

#include

int min(int a[10],int n)

{

int i,m=1;

for(i=2;i<=n;i++)

{

if(a[m]>a[i])

m=i;

}

return m;

}



void swap(int &a,int &b)

{

int t;

t=a;

a=b;

b=t;

}

int main()

{

int i,j,k,n,flag=0,l,m;

int d[100],a[100];

clrscr();

cout<<"\n\nenter n:\n\n";

cin>>n;

for(i=1;i<=n;i++)

a[i]=i;

for(i=1;i<=n;i++)

d[i]=i-1;

cout<<"\n\npermutations generated for integer 12...n

are:\n\n";

for(i=1;i<=n;i++) //display the given no.

cout<
cout<<"\t";

do

{

flag=0;

k=min(a,n);

for(i=1;i<=n;i++)

{

if((a[i]>a[k])&&(d[i]!=0)&&(a[d[i]]
mobile integer.

k=i;

}

if(a[k]==1)

break;

l=d[k]; //copy of direction of mobile integer.

m=a[k]; //copy of mobile integer.





if(d[k]==k+1&&d[k+1]==k) //swap directions.

{



if(d[k]==n)

{ d[k+1]=0;

d[k]=k-1;

}



else if(d[k+1]==1)

{

d[k]=0;

d[k+1]=k+2;

}

else

{ d[k]=k-1;

d[k+1]=k+2;

}

}

else if(d[k]==k-1&&d[k-1]==k)

{

d[k]=k+1;

d[k-1]=k-2;

}

/*cout<
cout<<"\t";

if u want to know directions of integers.

*/

swap(a[k],a[l]); //swap mobile integer and integer it is

pointing to.

for(i=1;i<=n;i++)

cout<
cout<<"\t";

for(i=1;i<=n;i++) //reverse directions of integers

greater than

//mobile integer.

{

if(a[i]>m)

{

if(d[i]==0&&i==n)

d[i]=i-1;

else if(d[i]
d[i]=i+1;

else if(d[i]>i)

d[i]=i-1;

}

}

for(i=1;i
{

if(a[i]
{



if(d[i+1]!=0)

flag=1;

}



else if(a[i]>a[i+1])

{



if(d[i]!=0)

flag=1;

}

}

}

while(flag==1); //if no mobile

integer(flag=0)terminate the program.

getch();

return 0;

}

Answer 2

Abcd abdc acbd acdb adbc adcb



And so forth switching the first letter

Answer 3

Can you add what you've done so far?