Combinations in a character array.

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Write a program that takes input a char array and outputs all the combinations of the characters in
By Abraham on Thursday, May 29, 2008 (UMST)

/*
Write a program that takes input a char array and outputs all the combinations of the characters in the character array. Example: consier char array 'a','b','c'} the output shouold be abc,cab,bac,acb,cba,bca that is all the combinations of characters 'a','b','c'.
*/

/*
AG: concatinate string abc to abcabc
the parse till len of given string
abc
bca
cab
now start from reverse (len*2) - 1 len times
cba
bcc
acb
*/

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

main()
{
char a[3] = {'a','b','c'};
char buffer[100] = {'\0'};
int len = sizeof(a)/sizeof(char);
int i =0, j = 0;
memset (buffer,'\0',100);
for (i = 0 ;i < len ;i++)
{
  buffer[i] = a[i];
}
strcat(buffer,buffer);
printf ("buffer = %s\n",buffer);

for (i = 0 ;i<len;i++)
{
  for (j =0;j<len;j++)
  {
   printf ("%c",buffer[i+j]);
  }
  printf ("\n");
}

for (i = (len*2) -1 ;i>= len ;i--)
{
  for (j =0;j<len;j++)
  {
   printf ("%c",buffer[i-j]);
  }
  printf ("\n");
}
}

/*

Output:
buffer = abcabc
abc
bca
cab
cba
bac
acb
Press any key to continue
*/

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doesn't work for more values
By ksumit on Saturday, June 14, 2008 (UMST)

Though this does work for {a,b,c}; it fails miserably for n>3
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Solution in C# (Works for n>3 also)
By kp.zeus on Saturday, July 05, 2008 (UMST)

public static string[] GetCombinations(char[] inputs) { if (inputs == null) { throw new ArgumentNullException("inputs"); } string[] combinations = new string[inputs.Length * (inputs.Length - 1)]; int i = 0, j = 0, pos = 0, resultCount = 0; char[] temp = new char[inputs.Length]; while (resultCount < combinations.Length) { while (i < inputs.Length - 2 && resultCount < combinations.Length) { pos = 0; //Forward while (pos <= i) { temp[pos] = inputs[pos]; pos++; } j = pos; while (j < inputs.Length) { temp[j] = inputs[j]; j++; } combinations[resultCount] = new string(temp); resultCount++; //Reverse j = inputs.Length - 1; while (pos < inputs.Length) { temp[pos] = inputs[j]; j--; pos++; } i++; combinations[resultCount] = new string(temp); resultCount++; } i = 0; char start = inputs[0]; //Shift while (i < (inputs.Length - 1)) { inputs[i] = inputs[i + 1]; i++; } inputs[inputs.Length - 1] = start; i = 0; } return combinations; }
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c# version. ok with long inputs
By factorme on Thursday, September 04, 2008 (UMST)

internal class Program {
        private static void Main(string[] args) {
            string input = Console.ReadLine();
            Console.WriteLine("---------result----------");
            PrintAllCombinations(input);
            Console.WriteLine("---------end----------");
            Console.ReadLine();
        }

        /// <summary>
        /// Print all possible combinations of the chars inside a string.
        /// </summary>
        /// <remarks>
        /// Algorithom:
        /// 1. Simulate the input with a number 1234...n where n is the input's length
        /// 2. Increase the number by 1 until the result exceeds n....321
        /// 3. Generate an result string based on each calculation result of the steps looped in 2.
        /// 2. Use an int array to simulate big numbers in case the input string is very long.
        /// </remarks>
        /// <param name="input"></param>
        private static void PrintAllCombinations(string input) {
            if (string.IsNullOrEmpty(input) || input.Length == 1) {
                Console.WriteLine(input);
                return;
            }

            int len = input.Length;
            int[] indice = new int[len];
            for (int i = 0; i < len; i++) {
                indice[i] = i;
            }

            bool goon = true;
            int toAdd = 0;
            while (goon) {
                Print(input, indice);
                int x = indice[len - 1];
                x += 1 + toAdd;
                if (x > len) {
                    indice[len - 1] = 0;
                    for (int i = len - 2; i >= 0; i--) {
                        int temp = indice[i];
                        temp++;
                        if (temp > len) {
                            if (i == 0) {
                                goon = false;
                            }

                            indice[i] = 0;
                        } else {
                            indice[i] = temp;
                            break;
                        }
                    }
                } else {
                    indice[len - 1] = x;
                }
            }
        }

        private static void Print(string input, params int[] indice) {
            for (int i = 0; i < input.Length; i++) {
                if (Array.IndexOf(indice, i) == -1) {
                    return;
                }
            }

            string temp = string.Empty;
            foreach (int i in indice ?? new int[0]) {
                temp += input[i];
            }
            Console.WriteLine(_count + ":\t" + temp);
            _count++;
        }

        private static int _count = 1;
    }

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This is very simple
By aravind on Thursday, October 09, 2008 (UMST)

//If the string is in letter and len is the length

string letter;

for (i=0;i<n;i++)

for(j=0;j<n;j++)

{

temp=letter[i];

letter[i+1]=letter[i];

letter[i+1]=temp;

cout<< letter;

}

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Simple answer
By aravind on Thursday, October 09, 2008 (UMST)

Sorry the above post had a mistake and i dont see option to delete

//If the string is in letter and len is the length

string letter;

for (i=0;i<n;i++)

for(j=0;j<n;j++)

{

temp=letter[j];

letter[j+1]=letter[j];

letter[j+1]=temp;

cout<< letter;

}

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Program in Java
By yogi_1306 on Saturday, October 11, 2008 (UMST)

public class Permutation { public static void main(String[] args){ permute("abcd"); } public static void permute(String letters) { permute(letters, 0); } private static String swap(String letters, int i, int j){ char[] asArray = letters.toCharArray(); char tmp = asArray[i]; asArray[i] = asArray[j]; asArray[j] = tmp; return new String(asArray); } private static void permute(String letters, int k) { if (k == letters.length()-1) { System.out.println(letters); return; } for (int idx = k; idx < letters.length(); idx++){ letters = swap(letters, k, idx); permute(letters, k+1); } } }
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