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:
The first step
All the other steps (all with the same logic)
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) -> a
If 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);
}
}
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.
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;
}
}
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;
}
}
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
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.
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;
}
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):
This: 36288000 items in 235 millisecs
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);
Base/Revise on Pengyang answer
And inspired from permutations-in-javascript
The c# version FunctionalPermutations should 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)));
}
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();
}
}
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);
}
}
}
}
I've got a two-dimensional array,
string[,] table = {
{ "aa", "aaa" },
{ "bb", "bbb" }
};
And I'd like to foreach through it like this,
foreach (string[] row in table)
{
Console.WriteLine(row[0] + " " + row[1]);
}
But, I get the error:
Can't convert type string to string[]
Is there a way I can achieve what I want, i.e. iterate through the first dimension of the array with the iterator variable returning me the one-dimensional array for that row?
Multidimensional arrays aren't enumerable. Just iterate the good old-fashioned way:
for (int i = 0; i < table.GetLength(0); i++)
{
Console.WriteLine(table[i, 0] + " " + table[i, 1]);
}
As others have suggested, you could use nested for-loops or redeclare your multidimensional array as a jagged one.
However, I think it's worth pointing out that multidimensional arrays are enumerable, just not in the way that you want. For example:
string[,] table = {
{ "aa", "aaa" },
{ "bb", "bbb" }
};
foreach (string s in table)
{
Console.WriteLine(s);
}
/* Output is:
aa
aaa
bb
bbb
*/
If you define your array like this:
string[][] table = new string[][] {
new string[] { "aa", "aaa" },
new string[]{ "bb", "bbb" }
};
Then you can use a foreach loop on it.
UPDATE: I had some time on my hands, so ... I went ahead and fleshed out this idea. See below for the code.
Here's a bit of a crazy answer:
You could do what you're looking for -- essentially treat a two-dimensional array as a table with rows -- by writing a static method (perhaps an extension method) that takes a T[,] and returns an IEnumerable<T[]>. This would require copying each "row" of the underlying table into a new array, though.
A perhaps better (though more involved) approach would be to actually write a class that implements IList<T> as a wrapper around a single "row" of a two-dimensional array (you would probably set IsReadOnly to true and just implement the getter for the this[int] property and probably Count and GetEnumerator; everything else could throw a NotSupportedException). Then your static/extension method could return an IEnumerable<IList<T>> and provide deferred execution.
That way you could write code pretty much like what you have:
foreach (IList<string> row in table.GetRows()) // or something
{
Console.WriteLine(row[0] + " " + row[1]);
}
Just a thought.
Implementation suggestion:
public static class ArrayTableHelper {
public static IEnumerable<IList<T>> GetRows<T>(this T[,] table) {
for (int i = 0; i < table.GetLength(0); ++i)
yield return new ArrayTableRow<T>(table, i);
}
private class ArrayTableRow<T> : IList<T> {
private readonly T[,] _table;
private readonly int _count;
private readonly int _rowIndex;
public ArrayTableRow(T[,] table, int rowIndex) {
if (table == null)
throw new ArgumentNullException("table");
if (rowIndex < 0 || rowIndex >= table.GetLength(0))
throw new ArgumentOutOfRangeException("rowIndex");
_table = table;
_count = _table.GetLength(1);
_rowIndex = rowIndex;
}
// I didn't implement the setter below,
// but you easily COULD (and then set IsReadOnly to false?)
public T this[int index] {
get { return _table[_rowIndex, index]; }
set { throw new NotImplementedException(); }
}
public int Count {
get { return _count; }
}
bool ICollection<T>.IsReadOnly {
get { return true; }
}
public IEnumerator<T> GetEnumerator() {
for (int i = 0; i < _count; ++i)
yield return this[i];
}
// omitted remaining IList<T> members for brevity;
// you actually could implement IndexOf, Contains, etc.
// quite easily, though
}
}
...now I think I should give StackOverflow a break for the rest of the day ;)
It depends on how you define your multi-dimensional array. Here are two options:
using System;
using System.Collections.Generic;
using System.Linq;
namespace ConsoleApplication2
{
class Program
{
static void Main(string[] args)
{
// First
string[,] arr1 = {
{ "aa", "aaa" },
{ "bb", "bbb" }
};
// Second
string[][] arr2 = new[] {
new[] { "aa", "aaa" },
new[] { "bb", "bbb" }
};
// Iterate through first
for (int x = 0; x <= arr1.GetUpperBound(0); x++)
for (int y = 0; y <= arr1.GetUpperBound(1); y++)
Console.Write(arr1[x, y] + "; ");
Console.WriteLine(Environment.NewLine);
// Iterate through second second
foreach (string[] entry in arr2)
foreach (string element in entry)
Console.Write(element + "; ");
Console.WriteLine(Environment.NewLine);
Console.WriteLine("Press any key to finish");
Console.ReadKey();
}
}
}
Here's a simple extension method that returns each row as an IEnumerable<T>. This has the advantage of not using any extra memory:
public static class Array2dExt
{
public static IEnumerable<IEnumerable<T>> Rows<T>(this T[,] array)
{
for (int r = array.GetLowerBound(0); r <= array.GetUpperBound(0); ++r)
yield return row(array, r);
}
static IEnumerable<T> row<T>(T[,] array, int r)
{
for (int c = array.GetLowerBound(1); c <= array.GetUpperBound(1); ++c)
yield return array[r, c];
}
}
Sample usage:
static void Main()
{
string[,] siblings = { { "Mike", "Amy" }, { "Mary", "Albert" }, {"Fred", "Harry"} };
foreach (var row in siblings.Rows())
Console.WriteLine("{" + string.Join(", ", row) + "}");
}
string[][] table = { ... };
string[][] languages = new string[2][];
languages[0] = new string[2];
languages[1] = new string[3];
// inserting data into double dimensional arrays.
for (int i = 0; i < 2; i++)
{
languages[0][i] = "Jagged"+i.ToString();
}
for (int j = 0; j < 3; j++)
{
languages[1][j] = "Jag"+j.ToString();
}
// doing foreach through 2 dimensional arrays.
foreach (string[] s in languages)
{
foreach (string a in s)
{
Console.WriteLine(a);
}
}
Using LINQ you can do it like this:
var table_enum = table
// Convert to IEnumerable<string>
.OfType<string>()
// Create anonymous type where Index1 and Index2
// reflect the indices of the 2-dim. array
.Select((_string, _index) => new {
Index1 = (_index / 2),
Index2 = (_index % 2), // ← I added this only for completeness
Value = _string
})
// Group by Index1, which generates IEnmurable<string> for all Index1 values
.GroupBy(v => v.Index1)
// Convert all Groups of anonymous type to String-Arrays
.Select(group => group.Select(v => v.Value).ToArray());
// Now you can use the foreach-Loop as you planned
foreach(string[] str_arr in table_enum) {
// …
}
This way it is also possible to use the foreach for looping through the columns instead of the rows by using Index2 in the GroupBy instead of Index 1. If you don't know the dimension of your array then you have to use the GetLength() method to determine the dimension and use that value in the quotient.
I'm not a big fan of this method because of the memory usage involved, but if you use the arrays it produces, it isn't such a waste.
public static void ForEachRow<T>(this T[,] list, Action<int, T[]> action)
{
var len = list.GetLength(0);
var sub = list.GetLength(1);
T[] e;
int i, j;
for (i = 0; i < len; i++)
{
e = new T[sub];
for (j = 0; j < sub; j++)
{
e[j] = list[i, j];
}
action(i, e);
}
}
Implementation:
var list = new[,]{0x0, 0x1, 0x2, 0x4, 0x8};
list.ForEachRow((i, row) =>
{
for (var j = 0; j < row.Length; j++)
{
Console.WriteLine("[{0},{1}]: {2}", i, j, row[j]);
}
});
The other solution I found is less memory intensive, but will use more CPU, especially when the dimensions of the arrays' entries are larger.
public static void ForEachRow<T>(this T[,] list, Action<int, IEnumerable<T>> action)
{
var len = list.GetLength(0);
var sub = list.GetLength(1);
int i, j;
IEnumerable<T> e;
for (i = 0; i < len; i++)
{
e = Enumerable.Empty<T>();
for (j = 0; j < sub; j++)
{
e = e.Concat(AsEnumerable(list[i, j]));
}
action(i, e);
}
}
private static IEnumerable<T> AsEnumerable<T>(T add)
{
yield return add;
}
Implementation:
var list = new[,]{0x0, 0x1, 0x2, 0x4, 0x8};
list.ForEachRow((i, row) =>
{
var j = 0;
forrach (var o in row)
{
Console.WriteLine("[{0},{1}]: {2}", i, j, o);
++j;
}
});
As a whole, I find the first option to be more intuitive, especially if you want to access the produced array by its indexer.
At the end of the day, this is all just eye candy, neither methods should really be used in favour of directly accessing the source array;
for (var i = 0; i < list.GetLength(0); i++)
{
foreach (var j = 0; j < list.GetLength(1); j++)
{
Console.WriteLine("[{0},{1}]: {2}", i, j, list[i, j]);
}
}
Remember that a multi-dimensional array is like a table. You don't have an x element and a y element for each entry; you have a string at (for instance) table[1,2].
So, each entry is still only one string (in your example), it's just an entry at a specific x/y value. So, to get both entries at table[1, x], you'd do a nested for loop. Something like the following (not tested, but should be close)
for (int x = 0; x < table.Length; x++)
{
for (int y = 0; y < table.Length; y += 2)
{
Console.WriteLine("{0} {1}", table[x, y], table[x, y + 1]);
}
}
I try this. I hope to help. It work with
static void Main()
{
string[,] matrix = {
{ "aa", "aaa" },
{ "bb", "bbb" }
};
int index = 0;
foreach (string element in matrix)
{
if (index < matrix.GetLength(1))
{
Console.Write(element);
if (index < (matrix.GetLength(1) - 1))
{
Console.Write(" ");
}
index++;
}
if (index == matrix.GetLength(1))
{
Console.Write("\n");
index = 0;
}
}