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 an example of how this is done and the logic behind solving such a problem?
I've seen a few code snippets but they weren't well commented/explained and thus hard to follow.
First of all: it smells like recursion of course!
Since you also wanted to know the principle, I did my best to explain it human language. I think recursion is very easy most of the times. You only have to grasp two steps:
In human language:
In short:
- The permutation of 1 element is one element.
- The permutation of a set of elements is a list each of the elements, concatenated with every permutation of the other elements.
Example:
If the set just has one element -->
return it.
perm(a) -> aIf the set has two characters: for each element in it: return the element, with the permutation of the rest of the elements added, like so:
perm(ab) ->
a + perm(b) -> ab
b + perm(a) -> ba
Further: for each character in the set: return a character, concatenated with a permutation of > the rest of the set
perm(abc) ->
a + perm(bc) --> abc, acb
b + perm(ac) --> bac, bca
c + perm(ab) --> cab, cba
perm(abc...z) -->
a + perm(...), b + perm(....)
....
I found the pseudocode on http://www.programmersheaven.com/mb/Algorithms/369713/369713/permutation-algorithm-help/:
makePermutations(permutation) {
if (length permutation < required length) {
for (i = min digit to max digit) {
if (i not in permutation) {
makePermutations(permutation+i)
}
}
}
else {
add permutation to list
}
}
C#
OK, and something more elaborate (and since it is tagged c #), from http://radio.weblogs.com/0111551/stories/2002/10/14/permutations.html : Rather lengthy, but I decided to copy it anyway, so the post is not dependent on the original.
The function takes a string of characters, and writes down every possible permutation of that exact string, so for example, if "ABC" has been supplied, should spill out:
ABC, ACB, BAC, BCA, CAB, CBA.
Code:
class Program
{
private static void Swap(ref char a, ref char b)
{
if (a == b) return;
var temp = a;
a = b;
b = temp;
}
public static void GetPer(char[] list)
{
int x = list.Length - 1;
GetPer(list, 0, x);
}
private static void GetPer(char[] list, int k, int m)
{
if (k == m)
{
Console.Write(list);
}
else
for (int i = k; i <= m; i++)
{
Swap(ref list[k], ref list[i]);
GetPer(list, k + 1, m);
Swap(ref list[k], ref list[i]);
}
}
static void Main()
{
string str = "sagiv";
char[] arr = str.ToCharArray();
GetPer(arr);
}
}
Swap(ref list[k], ref list[i]);) is unncessary. –
Wingding (a[x], a[y]) = (a[y], a[x]); –
Comparison It's just two lines of code if LINQ is allowed to use. Please see my answer here.
EDIT
Here is my generic function which can return all the permutations (not combinations) from a list of T:
static IEnumerable<IEnumerable<T>>
GetPermutations<T>(IEnumerable<T> list, int length)
{
if (length == 1) return list.Select(t => new T[] { t });
return GetPermutations(list, length - 1)
.SelectMany(t => list.Where(e => !t.Contains(e)),
(t1, t2) => t1.Concat(new T[] { t2 }));
}
Example:
IEnumerable<IEnumerable<int>> result =
GetPermutations(Enumerable.Range(1, 3), 3);
Output - a list of integer-lists:
{1,2,3} {1,3,2} {2,1,3} {2,3,1} {3,1,2} {3,2,1}
As this function uses LINQ so it requires .net 3.5 or higher.
const string s = "HALLOWEEN"; var result = GetPermutations(Enumerable.Range(0, s.Length), s.Length).Select(t => t.Select(i => s[i])); –
Curvaceous E, perm 2 selects the second E, etc) meaning duplicates in the output –
Hydrobomb var watch = new Stopwatch(); watch.Start(); for (int i = 0; i < 10; i++) GeneratePermutations(Enumerable.Range(1, 8).ToArray(), 0).ToList(); watch.Stop(); watch.ElapsedMilliseconds –
Atticism .ToArray() on it. In Pengyang's version you have to do .Select(v => v.ToArray()).ToArray() instead, but it does work as intended. –
Hagioscope Here I have found the solution. It was written in Java, but I have converted it to C#. I hope it will help you.
Here's the code in C#:
static void Main(string[] args)
{
string str = "ABC";
char[] charArry = str.ToCharArray();
Permute(charArry, 0, 2);
Console.ReadKey();
}
static void Permute(char[] arry, int i, int n)
{
int j;
if (i==n)
Console.WriteLine(arry);
else
{
for(j = i; j <=n; j++)
{
Swap(ref arry[i],ref arry[j]);
Permute(arry,i+1,n);
Swap(ref arry[i], ref arry[j]); //backtrack
}
}
}
static void Swap(ref char a, ref char b)
{
char tmp;
tmp = a;
a=b;
b = tmp;
}
Recursion is not necessary, here is good information about this solution.
var values1 = new[] { 1, 2, 3, 4, 5 };
foreach (var permutation in values1.GetPermutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
var values2 = new[] { 'a', 'b', 'c', 'd', 'e' };
foreach (var permutation in values2.GetPermutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
Console.ReadLine();
I have been used this algorithm for years, it has O(N) time and space complexity to calculate each permutation.
public static class SomeExtensions
{
public static IEnumerable<IEnumerable<T>> GetPermutations<T>(this IEnumerable<T> enumerable)
{
var array = enumerable as T[] ?? enumerable.ToArray();
var factorials = Enumerable.Range(0, array.Length + 1)
.Select(Factorial)
.ToArray();
for (var i = 0L; i < factorials[array.Length]; i++)
{
var sequence = GenerateSequence(i, array.Length - 1, factorials);
yield return GeneratePermutation(array, sequence);
}
}
private static IEnumerable<T> GeneratePermutation<T>(T[] array, IReadOnlyList<int> sequence)
{
var clone = (T[]) array.Clone();
for (int i = 0; i < clone.Length - 1; i++)
{
Swap(ref clone[i], ref clone[i + sequence[i]]);
}
return clone;
}
private static int[] GenerateSequence(long number, int size, IReadOnlyList<long> factorials)
{
var sequence = new int[size];
for (var j = 0; j < sequence.Length; j++)
{
var facto = factorials[sequence.Length - j];
sequence[j] = (int)(number / facto);
number = (int)(number % facto);
}
return sequence;
}
static void Swap<T>(ref T a, ref T b)
{
T temp = a;
a = b;
b = temp;
}
private static long Factorial(int n)
{
long result = n;
for (int i = 1; i < n; i++)
{
result = result * i;
}
return result;
}
}
O(N-1) for the sequence and O(N) for the swaps, which is O(N). And I'm still using this in production but with a refactor to generate only one permutation like: GetPermutation(i) where 0 <= i <= N!-1. I will be happy to use something with better performance than this, so be free to call a reference for something better, the most of the alternatives uses O(N!) in memory so you might check that too. –
Delft 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]);
}
}
}
Slightly modified version in C# that yields needed permutations in an array of ANY type.
// USAGE: create an array of any type, and call Permutations()
var vals = new[] {"a", "bb", "ccc"};
foreach (var v in Permutations(vals))
Console.WriteLine(string.Join(",", v)); // Print values separated by comma
public static IEnumerable<T[]> Permutations<T>(T[] values, int fromInd = 0)
{
if (fromInd + 1 == values.Length)
yield return values;
else
{
foreach (var v in Permutations(values, fromInd + 1))
yield return v;
for (var i = fromInd + 1; i < values.Length; i++)
{
SwapValues(values, fromInd, i);
foreach (var v in Permutations(values, fromInd + 1))
yield return v;
SwapValues(values, fromInd, i);
}
}
}
private static void SwapValues<T>(T[] values, int pos1, int pos2)
{
if (pos1 != pos2)
{
T tmp = values[pos1];
values[pos1] = values[pos2];
values[pos2] = tmp;
}
}
Permutations(vals).ToArray() then you end up with N references to the same array. If you want to be able to store the results you have to manually create a copy. E.g. Permutations(values).Select(v => (T[])v.Clone()) –
Hagioscope 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.
First of all, sets have permutations, not strings or integers, so I'll just assume you mean "the set of characters in a string."
Note that a set of size n has n! n-permutations.
The following pseudocode (from Wikipedia), called with k = 1...n! will give all the permutations:
function permutation(k, s) {
for j = 2 to length(s) {
swap s[(k mod j) + 1] with s[j]; // note that our array is indexed starting at 1
k := k / j; // integer division cuts off the remainder
}
return s;
}
Here's the equivalent Python code (for 0-based array indexes):
def permutation(k, s):
r = s[:]
for j in range(2, len(s)+1):
r[j-1], r[k%j] = r[k%j], r[j-1]
k = k/j+1
return r
k := k / j; does. –
Oubre Here is a simple solution in c# using recursion,
void Main()
{
string word = "abc";
WordPermuatation("",word);
}
void WordPermuatation(string prefix, string word)
{
int n = word.Length;
if (n == 0) { Console.WriteLine(prefix); }
else
{
for (int i = 0; i < n; i++)
{
WordPermuatation(prefix + word[i],word.Substring(0, i) + word.Substring(i + 1, n - (i+1)));
}
}
}
Here's a purely functional F# implementation:
let factorial i =
let rec fact n x =
match n with
| 0 -> 1
| 1 -> x
| _ -> fact (n-1) (x*n)
fact i 1
let swap (arr:'a array) i j = [| for k in 0..(arr.Length-1) -> if k = i then arr.[j] elif k = j then arr.[i] else arr.[k] |]
let rec permutation (k:int,j:int) (r:'a array) =
if j = (r.Length + 1) then r
else permutation (k/j+1, j+1) (swap r (j-1) (k%j))
let permutations (source:'a array) = seq { for k = 0 to (source |> Array.length |> factorial) - 1 do yield permutation (k,2) source }
Performance can be greatly improved by changing swap to take advantage of the mutable nature of CLR arrays, but this implementation is thread safe with regards to the source array and that may be desirable in some contexts. Also, for arrays with more than 16 elements int must be replaced with types with greater/arbitrary precision as factorial 17 results in an int32 overflow.
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
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]);
}
}
}
Here is the function which will print all permutaion. This function implements logic Explained by peter.
public class Permutation
{
//http://www.java2s.com/Tutorial/Java/0100__Class-Definition/RecursivemethodtofindallpermutationsofaString.htm
public static void permuteString(String beginningString, String endingString)
{
if (endingString.Length <= 1)
Console.WriteLine(beginningString + endingString);
else
for (int i = 0; i < endingString.Length; i++)
{
String newString = endingString.Substring(0, i) + endingString.Substring(i + 1);
permuteString(beginningString + endingString.ElementAt(i), newString);
}
}
}
static void Main(string[] args)
{
Permutation.permuteString(String.Empty, "abc");
Console.ReadLine();
}
The below is my implementation of permutation . Don't mind the variable names, as i was doing it for fun :)
class combinations
{
static void Main()
{
string choice = "y";
do
{
try
{
Console.WriteLine("Enter word :");
string abc = Console.ReadLine().ToString();
Console.WriteLine("Combinatins for word :");
List<string> final = comb(abc);
int count = 1;
foreach (string s in final)
{
Console.WriteLine("{0} --> {1}", count++, s);
}
Console.WriteLine("Do you wish to continue(y/n)?");
choice = Console.ReadLine().ToString();
}
catch (Exception exc)
{
Console.WriteLine(exc);
}
} while (choice == "y" || choice == "Y");
}
static string swap(string test)
{
return swap(0, 1, test);
}
static List<string> comb(string test)
{
List<string> sec = new List<string>();
List<string> first = new List<string>();
if (test.Length == 1) first.Add(test);
else if (test.Length == 2) { first.Add(test); first.Add(swap(test)); }
else if (test.Length > 2)
{
sec = generateWords(test);
foreach (string s in sec)
{
string init = s.Substring(0, 1);
string restOfbody = s.Substring(1, s.Length - 1);
List<string> third = comb(restOfbody);
foreach (string s1 in third)
{
if (!first.Contains(init + s1)) first.Add(init + s1);
}
}
}
return first;
}
static string ShiftBack(string abc)
{
char[] arr = abc.ToCharArray();
char temp = arr[0];
string wrd = string.Empty;
for (int i = 1; i < arr.Length; i++)
{
wrd += arr[i];
}
wrd += temp;
return wrd;
}
static List<string> generateWords(string test)
{
List<string> final = new List<string>();
if (test.Length == 1)
final.Add(test);
else
{
final.Add(test);
string holdString = test;
while (final.Count < test.Length)
{
holdString = ShiftBack(holdString);
final.Add(holdString);
}
}
return final;
}
static string swap(int currentPosition, int targetPosition, string temp)
{
char[] arr = temp.ToCharArray();
char t = arr[currentPosition];
arr[currentPosition] = arr[targetPosition];
arr[targetPosition] = t;
string word = string.Empty;
for (int i = 0; i < arr.Length; i++)
{
word += arr[i];
}
return word;
}
}
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;
}
And inspired from permutations-in-javascript
The c# version
FunctionalPermutationsshould be this
static IEnumerable<IEnumerable<T>> FunctionalPermutations<T>(IEnumerable<T> elements, int length)
{
if (length < 2) return elements.Select(t => new T[] { t });
/* Pengyang answser..
return _recur_(list, length - 1).SelectMany(t => list.Where(e => !t.Contains(e)),(t1, t2) => t1.Concat(new T[] { t2 }));
*/
return elements.SelectMany((element_i, i) =>
FunctionalPermutations(elements.Take(i).Concat(elements.Skip(i + 1)), length - 1)
.Select(sub_ei => new[] { element_i }.Concat(sub_ei)));
}
This is my solution which it is easy for me to understand
class ClassicPermutationProblem
{
ClassicPermutationProblem() { }
private static void PopulatePosition<T>(List<List<T>> finalList, List<T> list, List<T> temp, int position)
{
foreach (T element in list)
{
List<T> currentTemp = temp.ToList();
if (!currentTemp.Contains(element))
currentTemp.Add(element);
else
continue;
if (position == list.Count)
finalList.Add(currentTemp);
else
PopulatePosition(finalList, list, currentTemp, position + 1);
}
}
public static List<List<int>> GetPermutations(List<int> list)
{
List<List<int>> results = new List<List<int>>();
PopulatePosition(results, list, new List<int>(), 1);
return results;
}
}
static void Main(string[] args)
{
List<List<int>> results = ClassicPermutationProblem.GetPermutations(new List<int>() { 1, 2, 3 });
}
If performance and memory is an issue, I suggest this very efficient implementation. According to Heap's algorithm in Wikipedia, it should be the fastest. Hope it will fits your need :-) !
Just as comparison of this with a Linq implementation for 10! (code included):
Linq: 36288000 items in 50051 millisecs
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Runtime.CompilerServices;
using System.Text;
namespace WpfPermutations
{
/// <summary>
/// EO: 2016-04-14
/// Generator of all permutations of an array of anything.
/// Base on Heap's Algorithm. See: https://en.wikipedia.org/wiki/Heap%27s_algorithm#cite_note-3
/// </summary>
public static class Permutations
{
/// <summary>
/// Heap's algorithm to find all pmermutations. Non recursive, more efficient.
/// </summary>
/// <param name="items">Items to permute in each possible ways</param>
/// <param name="funcExecuteAndTellIfShouldStop"></param>
/// <returns>Return true if cancelled</returns>
public static bool ForAllPermutation<T>(T[] items, Func<T[], bool> funcExecuteAndTellIfShouldStop)
{
int countOfItem = items.Length;
if (countOfItem <= 1)
{
return funcExecuteAndTellIfShouldStop(items);
}
var indexes = new int[countOfItem];
for (int i = 0; i < countOfItem; i++)
{
indexes[i] = 0;
}
if (funcExecuteAndTellIfShouldStop(items))
{
return true;
}
for (int i = 1; i < countOfItem;)
{
if (indexes[i] < i)
{ // On the web there is an implementation with a multiplication which should be less efficient.
if ((i & 1) == 1) // if (i % 2 == 1) ... more efficient ??? At least the same.
{
Swap(ref items[i], ref items[indexes[i]]);
}
else
{
Swap(ref items[i], ref items[0]);
}
if (funcExecuteAndTellIfShouldStop(items))
{
return true;
}
indexes[i]++;
i = 1;
}
else
{
indexes[i++] = 0;
}
}
return false;
}
/// <summary>
/// This function is to show a linq way but is far less efficient
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="list"></param>
/// <param name="length"></param>
/// <returns></returns>
static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> list, int length)
{
if (length == 1) return list.Select(t => new T[] { t });
return GetPermutations(list, length - 1)
.SelectMany(t => list.Where(e => !t.Contains(e)),
(t1, t2) => t1.Concat(new T[] { t2 }));
}
/// <summary>
/// Swap 2 elements of same type
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="a"></param>
/// <param name="b"></param>
[MethodImpl(MethodImplOptions.AggressiveInlining)]
static void Swap<T>(ref T a, ref T b)
{
T temp = a;
a = b;
b = temp;
}
/// <summary>
/// Func to show how to call. It does a little test for an array of 4 items.
/// </summary>
public static void Test()
{
ForAllPermutation("123".ToCharArray(), (vals) =>
{
Debug.Print(String.Join("", vals));
return false;
});
int[] values = new int[] { 0, 1, 2, 4 };
Debug.Print("Non Linq");
ForAllPermutation(values, (vals) =>
{
Debug.Print(String.Join("", vals));
return false;
});
Debug.Print("Linq");
foreach(var v in GetPermutations(values, values.Length))
{
Debug.Print(String.Join("", v));
}
// Performance
int count = 0;
values = new int[10];
for(int n = 0; n < values.Length; n++)
{
values[n] = n;
}
Stopwatch stopWatch = new Stopwatch();
stopWatch.Reset();
stopWatch.Start();
ForAllPermutation(values, (vals) =>
{
foreach(var v in vals)
{
count++;
}
return false;
});
stopWatch.Stop();
Debug.Print($"Non Linq {count} items in {stopWatch.ElapsedMilliseconds} millisecs");
count = 0;
stopWatch.Reset();
stopWatch.Start();
foreach (var vals in GetPermutations(values, values.Length))
{
foreach (var v in vals)
{
count++;
}
}
stopWatch.Stop();
Debug.Print($"Linq {count} items in {stopWatch.ElapsedMilliseconds} millisecs");
}
}
}
Here's my solution in JavaScript (NodeJS). The main idea is that we take one element at a time, "remove it" from the string, vary the rest of the characters, and insert the element at the front.
function perms (string) {
if (string.length == 0) {
return [];
}
if (string.length == 1) {
return [string];
}
var list = [];
for(var i = 0; i < string.length; i++) {
var invariant = string[i];
var rest = string.substr(0, i) + string.substr(i + 1);
var newPerms = perms(rest);
for (var j = 0; j < newPerms.length; j++) {
list.push(invariant + newPerms[j]);
}
}
return list;
}
module.exports = perms;
And here are the tests:
require('should');
var permutations = require('../src/perms');
describe('permutations', function () {
it('should permute ""', function () {
permutations('').should.eql([]);
})
it('should permute "1"', function () {
permutations('1').should.eql(['1']);
})
it('should permute "12"', function () {
permutations('12').should.eql(['12', '21']);
})
it('should permute "123"', function () {
var expected = ['123', '132', '321', '213', '231', '312'];
var actual = permutations('123');
expected.forEach(function (e) {
actual.should.containEql(e);
})
})
it('should permute "1234"', function () {
// Wolfram Alpha FTW!
var expected = ['1234', '1243', '1324', '1342', '1423', '1432', '2134', '2143', '2314', '2341', '2413', '2431', '3124', '3142', '3214', '3241', '3412', '3421', '4123', '4132'];
var actual = permutations('1234');
expected.forEach(function (e) {
actual.should.containEql(e);
})
})
})
Here is the simplest solution I can think of:
let rec distribute e = function
| [] -> [[e]]
| x::xs' as xs -> (e::xs)::[for xs in distribute e xs' -> x::xs]
let permute xs = Seq.fold (fun ps x -> List.collect (distribute x) ps) [[]] xs
The distribute function takes a new element e and an n-element list and returns a list of n+1 lists each of which has e inserted at a different place. For example, inserting 10 at each of the four possible places in the list [1;2;3]:
> distribute 10 [1..3];;
val it : int list list =
[[10; 1; 2; 3]; [1; 10; 2; 3]; [1; 2; 10; 3]; [1; 2; 3; 10]]
The permute function folds over each element in turn distributing over the permutations accumulated so far, culminating in all permutations. For example, the 6 permutations of the list [1;2;3]:
> permute [1;2;3];;
val it : int list list =
[[3; 2; 1]; [2; 3; 1]; [2; 1; 3]; [3; 1; 2]; [1; 3; 2]; [1; 2; 3]]
Changing the fold to a scan in order to keep the intermediate accumulators sheds some light on how the permutations are generated an element at a time:
> Seq.scan (fun ps x -> List.collect (distribute x) ps) [[]] [1..3];;
val it : seq<int list list> =
seq
[[[]]; [[1]]; [[2; 1]; [1; 2]];
[[3; 2; 1]; [2; 3; 1]; [2; 1; 3]; [3; 1; 2]; [1; 3; 2]; [1; 2; 3]]]
Lists permutations of a string. Avoids duplication when characters are repeated:
using System;
using System.Collections;
class Permutation{
static IEnumerable Permutations(string word){
if (word == null || word.Length <= 1) {
yield return word;
yield break;
}
char firstChar = word[0];
foreach( string subPermute in Permutations (word.Substring (1)) ) {
int indexOfFirstChar = subPermute.IndexOf (firstChar);
if (indexOfFirstChar == -1) indexOfFirstChar = subPermute.Length;
for( int index = 0; index <= indexOfFirstChar; index++ )
yield return subPermute.Insert (index, new string (firstChar, 1));
}
}
static void Main(){
foreach( var permutation in Permutations ("aab") )
Console.WriteLine (permutation);
}
}
//Generic C# Method
private static List<T[]> GetPerms<T>(T[] input, int startIndex = 0)
{
var perms = new List<T[]>();
var l = input.Length - 1;
if (l == startIndex)
perms.Add(input);
else
{
for (int i = startIndex; i <= l; i++)
{
var copy = input.ToArray(); //make copy
var temp = copy[startIndex];
copy[startIndex] = copy[i];
copy[i] = temp;
perms.AddRange(GetPerms(copy, startIndex + 1));
}
}
return perms;
}
//usages
char[] charArray = new char[] { 'A', 'B', 'C' };
var charPerms = GetPerms(charArray);
string[] stringArray = new string[] { "Orange", "Mango", "Apple" };
var stringPerms = GetPerms(stringArray);
int[] intArray = new int[] { 1, 2, 3 };
var intPerms = GetPerms(intArray);
I hope this will suffice:
using System;
public class Program
{
public static void Main()
{
//Example using word cat
permute("cat");
}
static void permute(string word){
for(int i=0; i < word.Length; i++){
char start = word[0];
for(int j=1; j < word.Length; j++){
string left = word.Substring(1,j-1);
string right = word.Substring(j);
Console.WriteLine(start+right+left);
}
if(i+1 < word.Length){
word = wordChange(word, i + 1);
}
}
}
static string wordChange(string word, int index){
string newWord = "";
for(int i=0; i<word.Length; i++){
if(i== 0)
newWord += word[index];
else if(i== index)
newWord += word[0];
else
newWord += word[i];
}
return newWord;
}
output:
cat
cta
act
atc
tca
tac
Here is the function which will print all permutations recursively.
public void Permutations(string input, StringBuilder sb)
{
if (sb.Length == input.Length)
{
Console.WriteLine(sb.ToString());
return;
}
char[] inChar = input.ToCharArray();
for (int i = 0; i < input.Length; i++)
{
if (!sb.ToString().Contains(inChar[i]))
{
sb.Append(inChar[i]);
Permutations(input, sb);
RemoveChar(sb, inChar[i]);
}
}
}
private bool RemoveChar(StringBuilder input, char toRemove)
{
int index = input.ToString().IndexOf(toRemove);
if (index >= 0)
{
input.Remove(index, 1);
return true;
}
return false;
}
class Permutation
{
public static List<string> Permutate(string seed, List<string> lstsList)
{
loopCounter = 0;
// string s="\w{0,2}";
var lstStrs = PermuateRecursive(seed);
Trace.WriteLine("Loop counter :" + loopCounter);
return lstStrs;
}
// Recursive function to find permutation
private static List<string> PermuateRecursive(string seed)
{
List<string> lstStrs = new List<string>();
if (seed.Length > 2)
{
for (int i = 0; i < seed.Length; i++)
{
str = Swap(seed, 0, i);
PermuateRecursive(str.Substring(1, str.Length - 1)).ForEach(
s =>
{
lstStrs.Add(str[0] + s);
loopCounter++;
});
;
}
}
else
{
lstStrs.Add(seed);
lstStrs.Add(Swap(seed, 0, 1));
}
return lstStrs;
}
//Loop counter variable to count total number of loop execution in various functions
private static int loopCounter = 0;
//Non recursive version of permuation function
public static List<string> Permutate(string seed)
{
loopCounter = 0;
List<string> strList = new List<string>();
strList.Add(seed);
for (int i = 0; i < seed.Length; i++)
{
int count = strList.Count;
for (int j = i + 1; j < seed.Length; j++)
{
for (int k = 0; k < count; k++)
{
strList.Add(Swap(strList[k], i, j));
loopCounter++;
}
}
}
Trace.WriteLine("Loop counter :" + loopCounter);
return strList;
}
private static string Swap(string seed, int p, int p2)
{
Char[] chars = seed.ToCharArray();
char temp = chars[p2];
chars[p2] = chars[p];
chars[p] = temp;
return new string(chars);
}
}
Here is a C# answer which is a little simplified.
public static void StringPermutationsDemo()
{
strBldr = new StringBuilder();
string result = Permute("ABCD".ToCharArray(), 0);
MessageBox.Show(result);
}
static string Permute(char[] elementsList, int startIndex)
{
if (startIndex == elementsList.Length)
{
foreach (char element in elementsList)
{
strBldr.Append(" " + element);
}
strBldr.AppendLine("");
}
else
{
for (int tempIndex = startIndex; tempIndex <= elementsList.Length - 1; tempIndex++)
{
Swap(ref elementsList[startIndex], ref elementsList[tempIndex]);
Permute(elementsList, (startIndex + 1));
Swap(ref elementsList[startIndex], ref elementsList[tempIndex]);
}
}
return strBldr.ToString();
}
static void Swap(ref char Char1, ref char Char2)
{
char tempElement = Char1;
Char1 = Char2;
Char2 = tempElement;
}
Output:
1 2 3
1 3 2
2 1 3
2 3 1
3 2 1
3 1 2
Here is one more implementation of the algo mentioned.
public class Program
{
public static void Main(string[] args)
{
string str = "abcefgh";
var astr = new Permutation().GenerateFor(str);
Console.WriteLine(astr.Length);
foreach(var a in astr)
{
Console.WriteLine(a);
}
//a.ForEach(Console.WriteLine);
}
}
class Permutation
{
public string[] GenerateFor(string s)
{
if(s.Length == 1)
{
return new []{s};
}
else if(s.Length == 2)
{
return new []{s[1].ToString()+s[0].ToString(),s[0].ToString()+s[1].ToString()};
}
var comb = new List<string>();
foreach(var c in s)
{
string cStr = c.ToString();
var sToProcess = s.Replace(cStr,"");
if (!string.IsNullOrEmpty(sToProcess) && sToProcess.Length>0)
{
var conCatStr = GenerateFor(sToProcess);
foreach(var a in conCatStr)
{
comb.Add(c.ToString()+a);
}
}
}
return comb.ToArray();
}
}
new Permutation().GenerateFor("aba") outputs string[4] { "ab", "baa", "baa", "ab" } –
Chrotoem Here's another appraoch that is slighly more generic.
void Main()
{
var perms = new Permutations<char>("abc");
perms.Generate();
}
class Permutations<T> {
private List<T> permutation = new List<T>();
HashSet<T> chosen;
int n;
List<T> sequence;
public Permutations(IEnumerable<T> sequence)
{
this.sequence = sequence.ToList();
this.n = this.sequence.Count;
chosen = new HashSet<T>();
}
public void Generate()
{
if(permutation.Count == n) {
Console.WriteLine(string.Join(",",permutation));
}
else
{
foreach(var elem in sequence)
{
if(chosen.Contains(elem)) continue;
chosen.Add(elem);
permutation.Add(elem);
Generate();
chosen.Remove(elem);
permutation.Remove(elem);
}
}
}
}
Here's an answer which has the following features:
Supports permutation WITH repetition/duplicates
Iterative solution (no recursion needed)
Short and concise, with no need of special math operators or casting
Uses IEnumerable, so you can keep your code neat and tidy
Works on ints, but you can easily use string arrays and correspond them appropriately
static IEnumerable<int[]> permute(int size)
{
int[] i = new int[size];
while (true) {
yield return i;
int index = 0;
while (index < size) {
i[index]++;
if (i[index] == size) {
i[index] = 0;
index++;
}
else break;
}
if (index == size) break;
}
}
// Example usage:
static void Main(string[] args)
{
foreach (int[] i in permute(3))
{
foreach (int j in i) Console.Write(j + " ");
Console.WriteLine();
}
}
Output:
0 0 0
1 0 0
2 0 0
0 1 0
1 1 0
2 1 0
0 2 0
1 2 0
2 2 0
0 0 1
1 0 1
2 0 1
0 1 1
1 1 1
2 1 1
0 2 1
1 2 1
2 2 1
0 0 2
1 0 2
2 0 2
0 1 2
1 1 2
2 1 2
0 2 2
1 2 2
2 2 2
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