Java Program to print distinct permutations of a string

Given a string str, the task is to print all the distinct permutations of str. 
A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. 
For instance, the words ‘bat’ and ‘tab’ represents two distinct permutation (or arrangements) of a similar three letter word.

Examples:

Input: str = “abbb” 
Output: [abbb, babb, bbab, bbba]

Input: str = “abc” 
Output: [abc, bac, bca, acb, cab, cba] 

Approach: Write a recursive function that will generate all the permutations of the string. Terminating condition will be when the passed string is empty, in that case the function will return an empty ArrayList. Before adding the generated string, just check if it has already been generated before to get the distinct permutations.

Below is the implementation of the above approach:

Java

// Java implementation of the approach
import java.util.ArrayList;

public class GFG {

    // Function that returns true if string s
    // is present in the Arraylist
    static boolean isPresent(String s, ArrayList<String> Res)
    {

        // If present then return true
        for (String str : Res) {

            if (str.equals(s))
                return true;
        }

        // Not present
        return false;
    }

    // Function to return an ArrayList containing
    // all the distinct permutations of the string
    static ArrayList<String> distinctPermute(String str)
    {

        // If string is empty
        if (str.length() == 0) {

            // Return an empty arraylist
            ArrayList<String> baseRes = new ArrayList<>();
            baseRes.add("");
            return baseRes;
        }

        // Take first character of str
        char ch = str.charAt(0);

        // Rest of the string after excluding
        // the first character
        String restStr = str.substring(1);

        // Recurvise call
        ArrayList<String> prevRes = distinctPermute(restStr);

        // Store the generated sequence into
        // the resultant Arraylist
        ArrayList<String> Res = new ArrayList<>();
        for (String s : prevRes) {
            for (int i = 0; i <= s.length(); i++) {
                String f = s.substring(0, i) + ch + s.substring(i);

                // If the generated string is not
                // already present in the Arraylist
                if (!isPresent(f, Res))

                    // Add the generated string to the Arraylist
                    Res.add(f);
            }
        }

        // Return the resultant arraylist
        return Res;
    }

    // Driver code
    public static void main(String[] args)
    {
        String s = "abbb";
        System.out.println(distinctPermute(s));
    }
}

Output

[abbb, babb, bbab, bbba]

Time Complexity: O(n*n!)
Auxiliary space: O(n)

Optimization : We can optimize above solution to use HashSet to store result strings in place of Res ArrayList.