C# Program to print all permutations of a given string
A permutation also called an “arrangement number” or “order,” is a rearrangement of the elements of an ordered list S into a one-to-one correspondence with S itself. A string of length n has n! permutation.
Source: Mathword(http://mathworld.wolfram.com/Permutation.html)
Below are the permutations of string ABC.
ABC ACB BAC BCA CBA CAB
Here is a solution that is used as a basis in backtracking.
C#
// C# program to print all permutations // of a given string. using System; class GFG { /* Permutation function @param str string to calculate permutation for @param l starting index @param r end index */ private static void permute(String str, int l, int r) { if (l == r) Console.WriteLine(str); else { for ( int i = l; i <= r; i++) { str = swap(str, l, i); permute(str, l + 1, r); str = swap(str, l, i); } } } /* Swap Characters at position @param a string value @param i position 1 @param j position 2 @return swapped string */ public static String swap(String a, int i, int j) { char temp; char [] charArray = a.ToCharArray(); temp = charArray[i] ; charArray[i] = charArray[j]; charArray[j] = temp; string s = new string (charArray); return s; } // Driver Code public static void Main() { String str = "ABC" ; int n = str.Length; permute(str, 0, n-1); } } // This code is contributed by mits |
Output:
ABC ACB BAC BCA CBA CAB
Algorithm Paradigm: Backtracking
Time Complexity: O(n*n!) Note that there are n! permutations and it requires O(n) time to print a permutation.
Auxiliary Space: O(r – l)
Note: The above solution prints duplicate permutations if there are repeating characters in the input string. Please see the below link for a solution that prints only distinct permutations even if there are duplicates in input.
Print all distinct permutations of a given string with duplicates.
Permutations of a given string using STL
Another approach:
C#
// C# program to implement // the above approach using System; public class GFG{ static void permute(String s, String answer) { if (s.Length == 0) { Console.Write(answer + " " ); return ; } for ( int i = 0 ;i < s.Length; i++) { char ch = s[i]; String left_substr = s.Substring(0, i); String right_substr = s.Substring(i + 1); String rest = left_substr + right_substr; permute(rest, answer + ch); } } // Driver code public static void Main(String []args) { String s; String answer= "" ; Console.Write( "Enter the string : " ); s = Console.ReadLine(); Console.Write( "\nAll possible strings are : " ); permute(s, answer); } } // This code is contributed by gauravrajput1 |
Output:
Enter the string : abc All possible strings are : abc acb bac bca cab cba
Time Complexity: O(n*n!) The time complexity is the same as the above approach, i.e. there are n! permutations and it requires O(n) time to print a permutation.
Auxiliary Space: O(|s|)
Please refer complete article on Write a program to print all permutations of a given string for more details!