Create checklist of all feasible permutations of a string

Just how would certainly I deal with creating a checklist of all feasible permutations of a string in between x and also y personalities in size, having a variable checklist of personalities.

Any kind of language would certainly function, yet it needs to be mobile.

2019-05-03 18:57:49
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Answers: 4

You are going to get a great deal of strings, that's without a doubt ...

Where, x and also y is just how you specify them and also r is the variety of personalities we are picking from - - if I am recognizing you appropriately. You need to most definitely create these as required and also not get careless and also claim, create a powerset and afterwards filter the size of strings.

The adhering to most definitely isn't the most effective means to create these, yet it's an intriguing apart, none - the - much less.

Knuth (quantity 4, fascicle 2, informs us that (s, t) - mix amounts s+1 points taken t at once with rep - - an (s, t) - mix is symbols made use of by Knuth that amounts to {t \choose {s+t} We can figure this out by first creating each (s, t) - mix in binary kind (so, of size (s+t)) and also counting the variety of 0's to the left of each 1.

10001000011101 - - > comes to be the permutation : 0, 3, 4, 4, 4, 1

2019-05-08 15:53:17

I simply whipped this up fast in Ruby :

def perms(x, y, possible_characters)
  all = [""]
  current_array = all.clone
  1.upto(y) { |iteration|
    next_array = []
    current_array.each { |string|
      possible_characters.each { |c|
        value = string + c
        next_array.insert next_array.length, value
        all.insert all.length, value
    current_array = next_array
  all.delete_if { |string| string.length < x }

You could check into language API for constructed in permutation type features, and also you could be able to write even more maximized code, yet if the numbers are all that high, I'm not exactly sure there is much of a means around having a great deal of outcomes.

Anyways, the suggestion behind the code is start with string of size 0, after that track all the strings of size Z where Z is the existing dimension in the model. After that, experience each string and also add each personality onto each string. Ultimately at the end, remove any kind of that were listed below the x limit and also return the outcome.

I really did not examine it with possibly useless input (void personality checklist, unusual values of x and also y, etc ).

2019-05-07 18:10:56

I'm not exactly sure why you would certainly intend to do this to begin with. The resulting set for any kind of reasonably huge values of x and also y will certainly be massive, and also will certainly expand greatly as x and/or y grow.

Allows claim your set of feasible personalities is the 26 lowercase letters of the alphabet, and also you ask your application to create all permutations where size = 5. Thinking you do not lack memory you'll get 11,881,376 (i.e. 26 to the power of 5 ) strings back. Bump that size approximately 6, and also you'll get 308,915,776 strings back. These numbers get shateringly huge, really promptly.

Below's a remedy I create in Java. You'll require to give 2 runtime debates (representing x and also y ). Enjoy.

public class GeneratePermutations {
    public static void main(String[] args) {
        int lower = Integer.parseInt(args[0]);
        int upper = Integer.parseInt(args[1]);

        if (upper < lower || upper == 0 || lower == 0) {

        for (int length = lower; length <= upper; length++) {
            generate(length, "");

    private static void generate(int length, String partial) {
        if (length <= 0) {
        } else {
            for (char c = 'a'; c <= 'z'; c++) {
                generate(length - 1, partial + c);
2019-05-07 18:03:39

There are numerous means to do this. Usual approaches make use of recursion, memoization, or vibrant shows. The keynote is that you generate a checklist of all strings of size 1, after that in each model, for all strings generated in the last model, add that string concatenated with each personality in the string independently. (the variable index in the code listed below tracks the start of the last and also the next model )

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 would certainly after that require to remove all strings much less than x in size, they'll be the first (x-1 ) * len (originalString ) access in the checklist.

2019-05-07 16:58:26