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 = “cd”
Output: cd dc
Input: str = “abb”
Output: abb abb bab bba bab bba
Approach: Write a recursive function that prints every permutation of the given string. Termination condition will be when the passed string is empty.
Below is the implementation of the above approach:
:
public class CloudTechtwitter { static void printPermutn(String str, String ans) { if (str.length() == 0) { System.out.print(ans + " "); return; } for (int i = 0; i < str.length(); i++) { // ith character of str char ch = str.charAt(i); // Rest of the string after excluding // the ith character String ros = str.substring(0, i) + str.substring(i + 1); // Recursive call printPermutn(ros, ans + ch); } } // Driver code public static void main(String[] args) { String s = "abb"; printPermutn(s, ""); } }
abb abb bab bba bab bba
When the permutations need to be distinct.
Examples:
Input: str = “abb”
Output: abb bab bba
Approach: Write a recursive function that prints distinct permutations. Make a boolean array of size ’26’ which accounts for the character being used. If the character has not been used then the recursive call will take place.
Otherwise, don’t make any calls. The termination condition will be when the passed string is empty.
Below is the implementation of the above approach:
public class CloudTechTwitter { // Function to print all distinct permutations of str static void printDistinctPermutations(String str, String ans) { if (str.isEmpty()) { System.out.print(ans + " "); return; } boolean[] used = new boolean[26]; // To track used characters for (int i = 0; i < str.length(); i++) { char ch = str.charAt(i); String remaining = str.substring(0, i) + str.substring(i + 1); if (!used[ch - 'a']) { printDistinctPermutations(remaining, ans + ch); used[ch - 'a'] = true; } } } public static void main(String[] args) { String s = "abba"; printDistinctPermutations(s, ""); } }
without recursion
import java.util.HashSet; import java.util.LinkedList; import java.util.Queue; import java.util.Set; public class Permutations { // Function to print all distinct permutations of str static void printDistinctPermutations(String str) { if (str == null || str.length() == 0) { return; } // Queue to store permutations Queue<String> permutations = new LinkedList<>(); // Set to store distinct permutations Set<String> distinctPermutations = new HashSet<>(); // Initialize the queue with the first character permutations.add(""); // Process each character of the string for (char currentChar : str.toCharArray()) { int size = permutations.size(); // Generate new permutations by inserting the current character // into all positions of existing permutations while (size-- > 0) { String perm = permutations.poll(); for (int k = 0; k <= perm.length(); k++) { String newPerm = perm.substring(0, k) + currentChar + perm.substring(k); if (newPerm.length() == str.length()) { distinctPermutations.add(newPerm); } else { permutations.add(newPerm); } } } } // Print all distinct permutations for (String perm : distinctPermutations) { System.out.print(perm + " "); } } public static void main(String[] args) { String s = "abba"; printDistinctPermutations(s); } }