String Sort [codingblocks]

Nishant is a very naughty boy in Launchpad Batch. One day he was playing with strings, and randomly shuffled them all. Your task is to help Nishant Sort all the strings ( lexicographically ) but if a string is present completely as a prefix in another string, then string with longer length should come first. Eg bat, batman are 2 strings and the string bat is present as a prefix in Batman - then sorted order should have - Batman, bat.

Input Format

N(integer) followed by N strings.

Constraints

N<=1000

Output Format

N lines each containing one string.

Sample Input

3

bat

apple

batman

Sample Output

apple

batman

bat

Explanation

In mathematics, the lexicographic or lexicographical order (also known as lexical order, dictionary order, alphabetical order or lexicographic(al) product) is a generalization of the way words are alphabetically ordered based on the alphabetical order of their component letters.

Java Code:

IDE code: https://ide.geeksforgeeks.org/bfCU88Hs7h

/*
Amit Kumar
03-12-2020
IDE code: https://ide.geeksforgeeks.org/bfCU88Hs7h
*/

import java.util.ArrayList;
import java.util.Comparator;
import java.util.Scanner;

class Name {
    String name;

    Name(String name) {
        this.name = name;
    }
}

class SortByName implements Comparator<Name> {

    @Override
    public int compare(Name name1, Name name2) {
        if (name1.name.contains(name2.name) || name2.name.contains(name1.name)) {
            return name2.name.compareTo(name1.name);
        }
        return name1.name.compareTo(name2.name);
    }
}

public class Array_StringSort {
    public static Scanner scan = new Scanner(System.in);
    public static void main(String[] args) {
        int num = scan.nextInt();
        ArrayList<Name> names = new ArrayList<>();
        while (num-- >0) {
            String getname = scan.next();
            Name name = new Name(getname);
            names.add(name);
        }

        names.sort(new SortByName());
        for (Name nameObj: names){
            System.out.println(nameObj.name);
        }
    }
}