Algorithm to generate all permutation by selecting

Question

I need to generate all permutation of a string with selecting some of the elements. Like if my string is "abc" output would be { a,b,c,ab,ba,ac,ca,bc,cb,abc,acb,bac,bca,cab,cba }.

I thought a basic algorithm in which I generate all possible combination of "abc" which are {a,b,c,ab,ac,bc,abc} and then permute all of them.

So is there any efficient permutation algorithm by which I can generate all possible permutation with varying size.

The code I wrote for this is :

    #include <iostream>
    #include <stdio.h>
    #include <stdlib.h>
    #include <map>
    using namespace std;

    int permuteCount = 1;


    int compare (const void * a, const void * b)
    {
      return ( *(char*)a - *(char*)b);
    }

    void permute(char *str, int start, int end)
    {
        // cout<<"before sort : "<<str;

        // cout<<"after sort : "<<str;
          do
         {
               cout<<permuteCount<<")"<<str<<endl;  
               permuteCount++;
         }while( next_permutation(str+start,str+end) );  
    }

void generateAllCombinations( char* str)
{
     int     n, k, i, j, c;
     n = strlen(str);

     map<string,int> combinationMap;

for( k =1; k<=n; k++)
{  
   char tempStr[20];
   int index =0;
   for (i=0; i<(1<<n); i++) {
        index =0;
        for (j=0,c=0; j<32; j++) if (i & (1<<j)) c++;
        if (c == k) {

        for (j=0;j<32; j++) 
            if (i & (1<<j)) 
               tempStr[ index++] = str[j];          
        tempStr[index] = '\0';
        qsort (tempStr, index, sizeof(char), compare);
        if( combinationMap.find(tempStr) == combinationMap.end() )
        {
        //  cout<<"comb : "<<tempStr<<endl;
        //cout<<"unique comb : \n";
            combinationMap[tempStr] = 1; 
            permute(tempStr,0,k);   
        }  /*
        else
        {
            cout<<"duplicated comb : "<<tempStr<<endl;
        }*/
        }
  }


}
}


    int main () {


            char str[20];
            cin>>str;

            generateAllCombinations(str);

           cin>>str;
    }

I need to use a hash for avoiding same combination, so please let me know how can I make this algorithm better.

Thanks, GG

Answer 1

#include <algorithm>
#include <iostream>
#include <string>

int main() {
  using namespace std;
  string s = "abc";
  do {
    cout << s << '\n'; 
  } while (next_permutation(s.begin(), s.end()));
  return 0;
}

Next_permutation uses a constant size, but you can add a loop to deal with varying size. Or just store in a set to eliminate the extra dupes for you:

#include <set>

int main() {
  using namespace std;
  string s = "abc";
  set<string> results;
  do {
    for (int n = 1; n <= s.size(); ++n) {
      results.insert(s.substr(0, n));
    }
  } while (next_permutation(s.begin(), s.end()));
  for (set<string>::const_iterator x = results.begin(); x != results.end(); ++x) {
    cout << *x << '\n';
  }
  return 0;
}

Answer 2

I don't think you can write much faster program than you have already. The main problem is the output size: it has order of n!*2^n (number of subsets * average number of permutations for one subset), which is already > 10^9 for a string of 10 different characters.

Since STL's next_permutation adds very limited complexity for such small strings, your program's time complexity is already nearly O(output size).

But you can make your program a bit simpler. In particular, for( k =1; k<=n; k++) loop seems unnecessary: you already calculate size of subset in variable c inside. So, just have int k = c instead of if (c == k). (You'll also need to consider case of empty subset: i == 0)

edit
Actually, there's only 9864100 outputs for n == 10 (not ~ 10^9). Still, my point remains the same: your program already wastes only "O(next_permutation)" time for each output, which is very, very little.