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.
// Java program to print all // permutations of a given string. public class Permutation
{ public static void main(String[] args)
{
String str = "ABC" ;
int n = str.length();
Permutation permutation =
new Permutation();
permutation.permute(str, 0 , n- 1 );
}
/* Permutation function @param str
string to calculate permutation
for @param l starting index
@param r end index */
private void permute(String str,
int l, int r)
{
if (l == r)
System.out.println(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 String swap(String a,
int i, int j)
{
char temp;
char [] charArray = a.toCharArray();
temp = charArray[i] ;
charArray[i] = charArray[j];
charArray[j] = temp;
return String.valueOf(charArray);
}
} // This code is contributed by Mihir Joshi |
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:
import java.util.*;
// Java program to implement // the above approach class GFG{
static void permute(String s,
String answer)
{ if (s.length() == 0 )
{
System.out.print(answer + " " );
return ;
}
for ( int i = 0 ;i < s.length(); i++)
{
char ch = s.charAt(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[])
{ Scanner scan = new Scanner(System.in);
String s;
String answer= "" ;
System.out.print(
"Enter the string : " );
s = scan.next();
System.out.print(
"\nAll possible strings are : " );
permute(s, answer);
} } // This code is contributed by adityapande88 |
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|)