Listing all permutations of a string/integer

Question

A common task in programming interviews (not from my experience of interviews though) is to take a string or an integer and list every possible permutation.

Is there

Answer 1

class Program
{
    public static void Main(string[] args)
    {
        Permutation("abc");
    }

    static void Permutation(string rest, string prefix = "")
    {
        if (string.IsNullOrEmpty(rest)) Console.WriteLine(prefix);

        // Each letter has a chance to be permutated
        for (int i = 0; i < rest.Length; i++)
        {                
            Permutation(rest.Remove(i, 1), prefix + rest[i]);
        }
    }
}

Answer 2

Here is an easy to understand permutaion function for both string and integer as input. With this you can even set your output length(which in normal case it is equal to input length)

String

    static ICollection<string> result;

    public static ICollection<string> GetAllPermutations(string str, int outputLength)
    {
        result = new List<string>();
        MakePermutations(str.ToCharArray(), string.Empty, outputLength);
        return result;
    }

    private static void MakePermutations(
       char[] possibleArray,//all chars extracted from input
       string permutation,
       int outputLength//the length of output)
    {
         if (permutation.Length < outputLength)
         {
             for (int i = 0; i < possibleArray.Length; i++)
             {
                 var tempList = possibleArray.ToList<char>();
                 tempList.RemoveAt(i);
                 MakePermutations(tempList.ToArray(), 
                      string.Concat(permutation, possibleArray[i]), outputLength);
             }
         }
         else if (!result.Contains(permutation))
            result.Add(permutation);
    }

and for Integer just change the caller method and MakePermutations() remains untouched:

    public static ICollection<int> GetAllPermutations(int input, int outputLength)
    {
        result = new List<string>();
        MakePermutations(input.ToString().ToCharArray(), string.Empty, outputLength);
        return result.Select(m => int.Parse(m)).ToList<int>();
    }

example 1: GetAllPermutations("abc",3); "abc" "acb" "bac" "bca" "cab" "cba"

example 2: GetAllPermutations("abcd",2); "ab" "ac" "ad" "ba" "bc" "bd" "ca" "cb" "cd" "da" "db" "dc"

example 3: GetAllPermutations(486,2); 48 46 84 86 64 68

Answer 3

Here's a high level example I wrote which illustrates the human language explanation Peter gave:

    public List<string> FindPermutations(string input)
    {
        if (input.Length == 1)
            return new List<string> { input };
        var perms = new List<string>();
        foreach (var c in input)
        {
            var others = input.Remove(input.IndexOf(c), 1);
            perms.AddRange(FindPermutations(others).Select(perm => c + perm));
        }
        return perms;
    }

Answer 4

Building on @Peter's solution, here's a version that declares a simple LINQ-style Permutations() extension method that works on any IEnumerable<T>.

Usage (on string characters example):

foreach (var permutation in "abc".Permutations())
{
    Console.WriteLine(string.Join(", ", permutation));
}

Outputs:

a, b, c
a, c, b
b, a, c
b, c, a
c, b, a
c, a, b

Or on any other collection type:

foreach (var permutation in (new[] { "Apples", "Oranges", "Pears"}).Permutations())
{
    Console.WriteLine(string.Join(", ", permutation));
}

Outputs:

Apples, Oranges, Pears
Apples, Pears, Oranges
Oranges, Apples, Pears
Oranges, Pears, Apples
Pears, Oranges, Apples
Pears, Apples, Oranges

using System;
using System.Collections.Generic;
using System.Linq;

public static class PermutationExtension
{
    public static IEnumerable<T[]> Permutations<T>(this IEnumerable<T> source)
    {
        var sourceArray = source.ToArray();
        var results = new List<T[]>();
        Permute(sourceArray, 0, sourceArray.Length - 1, results);
        return results;
    }

    private static void Swap<T>(ref T a, ref T b)
    {
        T tmp = a;
        a = b;
        b = tmp;
    }

    private static void Permute<T>(T[] elements, int recursionDepth, int maxDepth, ICollection<T[]> results)
    {
        if (recursionDepth == maxDepth)
        {
            results.Add(elements.ToArray());
            return;
        }

        for (var i = recursionDepth; i <= maxDepth; i++)
        {
            Swap(ref elements[recursionDepth], ref elements[i]);
            Permute(elements, recursionDepth + 1, maxDepth, results);
            Swap(ref elements[recursionDepth], ref elements[i]);
        }
    }
}

Answer 5

I liked FBryant87 approach since it's simple. Unfortunately, it does like many other "solutions" not offer all permutations or of e.g. an integer if it contains the same digit more than once. Take 656123 as an example. The line:

var tail = chars.Except(new List<char>(){c});

using Except will cause all occurrences to be removed, i.e. when c = 6, two digits are removed and we are left with e.g. 5123. Since none of the solutions I tried solved this, I decided to try and solve it myself by FBryant87's code as base. This is what I came up with:

private static List<string> FindPermutations(string set)
    {
        var output = new List<string>();
        if (set.Length == 1)
        {
            output.Add(set);
        }
        else
        {
            foreach (var c in set)
            {
                // Remove one occurrence of the char (not all)
                var tail = set.Remove(set.IndexOf(c), 1);
                foreach (var tailPerms in FindPermutations(tail))
                {
                    output.Add(c + tailPerms);
                }
            }
        }
        return output;
    }

I simply just remove the first found occurrence using .Remove and .IndexOf. Seems to work as intended for my usage at least. I'm sure it could be made cleverer.

One thing to note though: The resulting list may contain duplicates, so make sure you either make the method return e.g. a HashSet instead or remove the duplicates after the return using any method you like.