Big Sorting | Editorial by aksingh007

Apply any efficient sorting technique and Sort on the basis of length of the string and then by comparing eachother.
T(n)=O(nlogn)
```
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

class X implements Comparable<X>{
    String str;
    X(String s){
        this.str=s;
    }
    public int compareTo(X x){
        if((this.str).length()!=x.str.length())
            return ( (this.str).length()-x.str.length() );
        else{
            return this.str.compareTo(x.str);
        }
    }
}

public class Solution {

public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        ArrayList<X> arr=new ArrayList<X>();
       // String[] unsorted = new String[n];
        for(int i=0; i < n; i++){
            arr.add(new X(in.next()));
       }
        Collections.sort(arr);
        for(X x:arr)
            System.out.println(x.str);
        
        // your code goes here
    }
}

```

Big Sorting | Editorial by aksingh007

Download submission

All caught up!

View submission

Is your resume even shortlisted?