Permutations

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
package code;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedHashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

/**
 *
 * @author manth
 */
public class Code {

    /**
     * @param args the command line arguments
     */
    /*
    Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
    */
    public static void main(String[] args) {
       Scanner sc=new Scanner(System.in);
       String s=sc.nextLine();


       permutations(0,"abc");







       }
    public static void permutations(int depth,String input)
    {
        if(depth==input.length())
        {
            System.out.println(input);
            return;
        }
       // char input_arr[]=input.toCharArray();
        for(int i=depth;i<input.length();i++)
        {
            char input_arr[]=input.toCharArray();
            char tmp=input_arr[depth];
            input_arr[depth]=input_arr[i];
            input_arr[i]=tmp;

            String swaped=new String(input_arr);
            permutations(depth+1,swaped);
        }

    }










}