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Print all permutation of a string using ArrayList
  • Last Updated : 13 Jun, 2019

Given a string str, the task is to print all the 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 = “abc”
Output: abc acb bac bca cba cab

Input: str = “bat”
Output: bat bta abt atb tba tab

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.



Below is the implementation of the above approach:

Java




// Java implementation of the approach
import java.util.ArrayList;
  
public class GFG {
  
    // Utility function to print the contents
    // of the ArrayList
    static void printArrayList(ArrayList<String> arrL)
    {
        arrL.remove("");
        for (int i = 0; i < arrL.size(); i++)
            System.out.print(arrL.get(i) + " ");
    }
  
    // Function to returns the arraylist which contains
    // all the permutation of str
    public static ArrayList<String> getPermutation(String str)
    {
  
        // If string is empty
        if (str.length() == 0) {
  
            // Return an empty arraylist
            ArrayList<String> empty = new ArrayList<>();
            empty.add("");
            return empty;
        }
  
        // Take first character of str
        char ch = str.charAt(0);
  
        // Take sub-string starting from the
        // second character
        String subStr = str.substring(1);
  
        // Recurvise call
        ArrayList<String> prevResult = getPermutation(subStr);
  
        // Store the generated permutations
        // into the resultant arraylist
        ArrayList<String> Res = new ArrayList<>();
  
        for (String val : prevResult) {
            for (int i = 0; i <= val.length(); i++) {
                Res.add(val.substring(0, i) + ch + val.substring(i));
            }
        }
  
        // Return the resultant arraylist
        return Res;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String str = "abc";
        printArrayList(getPermutation(str));
    }
}

C#




// C# implementation of the approach 
using System.Collections.Generic;
using System;
  
class GFG 
  
    // Utility function to print the contents 
    // of the ArrayList 
    static void printArrayList(List<String> arrL) 
    
        arrL.Remove(""); 
        for (int i = 0; i < arrL.Count; i++) 
            Console.Write(arrL[i] + " "); 
    
  
    // Function to returns the arraylist which contains 
    // all the permutation of str 
    public static List<String> getPermutation(String str) 
    
  
        // If string is empty 
        if (str.Length == 0)
        
  
            // Return an empty arraylist 
            List<String> empty = new List<String>(); 
            empty.Add(""); 
            return empty; 
        
  
        // Take first character of str 
        char ch = str[0]; 
  
        // Take sub-string starting from the 
        // second character 
        String subStr = str.Substring(1); 
  
        // Recurvise call 
        List<String> prevResult = getPermutation(subStr); 
  
        // Store the generated permutations 
        // into the resultant arraylist 
        List<String> Res = new List<String>(); 
  
        foreach (String val in prevResult)
        
            for (int i = 0; i <= val.Length; i++) 
            
                Res.Add(val.Substring(0, i) + ch + val.Substring(i)); 
            
        
  
        // Return the resultant arraylist 
        return Res; 
    
  
    // Driver code 
    public static void Main(String[] args) 
    
        String str = "abc"
        printArrayList(getPermutation(str)); 
    
  
// This code has been contributed by 29AjayKumar
Output:
abc bac bca acb cab cba

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