Java String Compare

 Hackerrak Java String Compare By Java Tech Solutionz

Problem:

We define the following terms:

• Lexicographical Order, also known as alphabetic or dictionary order, orders characters as follows:
  

  For example, ball < cat, dog < dorm, Happy < happy, Zoo < ball.

• A substring of a string is a contiguous block of characters in the string. For example, the substrings of abc are a, b, c, ab, bc, and abc.

Given a string, , and an integer, , complete the function so that it finds the lexicographically smallest and largest substrings of length .

Function Description

Complete the getSmallestAndLargest function in the editor below.

getSmallestAndLargest has the following parameters:

• string s: a string
• int k: the length of the substrings to find

Returns

• string: the string ' + "\n" + ' where and are the two substrings

Input Format

The first line contains a string denoting .
The second line contains an integer denoting .

Constraints

•  consists of English alphabetic letters only (i.e., [a-zA-Z]).

Sample Input 0

welcometojava
3

Sample Output 0

ava
wel

Explanation 0

String  has the following lexicographically-ordered substrings of length :

We then return the first (lexicographically smallest) substring and the last (lexicographically largest) substring as two newline-separated values (i.e., ava\nwel).

The stub code in the editor then prints ava as our first line of output and wel as our second line of output.

Solution:

public class Solution {

    public static String getSmallestAndLargest(String s, int k) {

        String smallest = "";

        String largest = "";

        

        // Complete the function

        // 'smallest' must be the lexicographically smallest substring of length 'k'

        // 'largest' must be the lexicographically largest substring of length 'k'

        

      

        

        int n=0;

        for(int i=0 ; i<=s.length()-k;i++){

            n++;

        }

        String arr[] = new String[n];

        

        for(int i=0 ; i<=s.length()-k;i++){

          

            arr[i]=s.substring(i,i+k);

        }

        arr = mySort(arr);

        System.out.println(arr[0]);

        System.out.println(arr[n-1]);

        

        return smallest + "\n" + largest;

    }

    

    public static String[] mySort(String[] ar) {

        

        String temp;

        

        for(int i=ar.length-1; i>0 ; i--) {

            for(int j=0;j<i;j++) {

                if(ar[j].compareTo(ar[j+1] )>0) {

                    temp = ar[j];

                    ar[j] = ar[j+1];

                    ar[j+1] = temp;

                }

            }

        }

    

        return ar;

    }