Generating permutation of string without duplicates [duplicate] - c++

Question

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Finding all the unique permutations of a string without generating duplicates
(5 answers)
Closed 9 years ago.
I have writing a general program to generate permutation of string but removing the duplicate cases . For this I am using memorization by using .
void permute(char *a,int i, int n,set<char*> s)
{
if(i==n)
{
if(s.find(a)==s.end()){
cout<<"no dublicate"<<endl;
cout<<a<<endl;
s.insert(a)
}
}
else{
for(int j=i;j<n;j++)
{
swap(a[i],a[j]);
permute(a,i+1,n,s);
swap(a[i],a[j]);
}
}
}
int main()
{
char a[]="aba";
set <char*> s;
permute(a,0,3,s);
return 0;
}
But the result is not as desired. It prints all the permutation. Can anyone help me in figuring out the problem.

Answer 1

First, you pass set<> s parameter by value, which discards your each insert, because it's done in the local copy of s only. However even if you change it to pass by reference, it won't work, because every time you insert the same char* value, so only one insert will be done. To make your code work correctly I suggest to change the prototype of your function to
void permute(string a,int i, int n,set<string>& s)
and this works all right.
update: source code with described minor changes
void permute(string a,int i, int n,set<string>& s)
{
if(i==n)
{
if(s.find(a)==s.end()){
cout<<"no dublicate"<<endl;
cout<<a<<endl;
s.insert(a);
}
}
else{
for(int j=i;j<n;j++)
{
swap(a[i],a[j]);
permute(a,i+1,n,s);
swap(a[i],a[j]);
}
}
}
int main()
{
string a ="aba";
set <string> s;
permute(a,0,3,s);
return 0;
}