Generate list of all possible permutations of a string

How would I go about generating a list of all possible permutations of a string between x and y characters in length, containing a variable list of characters.

Any language would work, but it should be portable.

Solution 1:

There are several ways to do this. Common methods use recursion, memoization, or dynamic programming. The basic idea is that you produce a list of all strings of length 1, then in each iteration, for all strings produced in the last iteration, add that string concatenated with each character in the string individually. (the variable index in the code below keeps track of the start of the last and the next iteration)

Some pseudocode:

list = originalString.split('')
index = (0,0)
list = [""]
for iteration n in 1 to y:
  index = (index[1], len(list))
  for string s in list.subset(index[0] to end):
    for character c in originalString:
      list.add(s + c)

you'd then need to remove all strings less than x in length, they'll be the first (x-1) * len(originalString) entries in the list.

Solution 2:

It's better to use backtracking

#include <stdio.h>
#include <string.h>

void swap(char *a, char *b) {
    char temp;
    temp = *a;
    *a = *b;
    *b = temp;
}

void print(char *a, int i, int n) {
    int j;
    if(i == n) {
        printf("%s\n", a);
    } else {
        for(j = i; j <= n; j++) {
            swap(a + i, a + j);
            print(a, i + 1, n);
            swap(a + i, a + j);
        }
    }
}

int main(void) {
    char a[100];
    gets(a);
    print(a, 0, strlen(a) - 1);
    return 0;
}

Solution 3:

You are going to get a lot of strings, that's for sure...

Where x and y is how you define them and r is the number of characters we are selecting from --if I am understanding you correctly. You should definitely generate these as needed and not get sloppy and say, generate a powerset and then filter the length of strings.

The following definitely isn't the best way to generate these, but it's an interesting aside, none-the-less.

Knuth (volume 4, fascicle 2, 7.2.1.3) tells us that (s,t)-combination is equivalent to s+1 things taken t at a time with repetition -- an (s,t)-combination is notation used by Knuth that is equal to . We can figure this out by first generating each (s,t)-combination in binary form (so, of length (s+t)) and counting the number of 0's to the left of each 1.

10001000011101 --> becomes the permutation: {0, 3, 4, 4, 4, 1}

Solution 4:

Non recursive solution according to Knuth, Python example:

def nextPermutation(perm):
    k0 = None
    for i in range(len(perm)-1):
        if perm[i]<perm[i+1]:
            k0=i
    if k0 == None:
        return None

    l0 = k0+1
    for i in range(k0+1, len(perm)):
        if perm[k0] < perm[i]:
            l0 = i

    perm[k0], perm[l0] = perm[l0], perm[k0]
    perm[k0+1:] = reversed(perm[k0+1:])
    return perm

perm=list("12345")
while perm:
    print perm
    perm = nextPermutation(perm)