TLE in case 3 and rest cases are correct

#1

TLE in case 3 and rest cases are correct

#2

link to my code is https://ide.codingblocks.com/s/285168

#3

@jatin111 lemme check bro

#4

@jatin111 Bro don’t use inefficient sorting algorithm, instead just use Arrays.sort(name), it will pass all!

#5

Arrays.sort func does not allow char array

#6

@jatin111 First of all You are making Array of strings not char, recheck ya code.
Here is your corrected code, as I suggested.
import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
String str = sc.next();
ArrayList res = permutations(str, str.length());
int count = 1;
String t;
String name[] = res.toArray(new String[res.size()]);
Arrays.sort(name);
for (int i = 0; i < name.length; i++) {
if (name[i].equals(str)) {
break;
} else {
System.out.println(name[i]);
}
}
}

public static ArrayList<String> permutations(String str, int len) {
	if (str.length() == 0) {
		ArrayList<String> base = new ArrayList<>();
		base.add("");
		return base;
	}

	char ch = str.charAt(0);
	String ros = str.substring(1);

	ArrayList<String> rr = new ArrayList<>();
	ArrayList<String> mr = new ArrayList<>();

	rr = permutations(ros, len - 1);

	for (int i = 0; i < rr.size(); i++) {
		String s = rr.get(i);
		for (int j = 0; j <= s.length(); j++)
			mr.add(s.substring(0, j) + ch + s.substring(j, s.length()));
	}

	return mr;
}

}

#7

See this i have clearly mentioned name as array, and name stores strings!