Balanced String

Hi
Can you help solve this problem

Thanks

This piece of code will perform the desired task for us. If you find it little difficult to understand then we can make a little simpler code also which will also do the same.

import java.util.HashSet;
import java.util.Set;

public class findShortestBalancedSubstring {
	public static void main(String[] args) {
        System.out.println(getShortestFragment("ABcabbCa"));
        System.out.println(getShortestFragment("azABaabza"));
        System.out.println(getShortestFragment("CATattac"));
        System.out.println(getShortestFragment("TacoCat"));
        System.out.println(getShortestFragment("Madam"));
        System.out.println(getShortestFragment("AcZCbaBz"));
        System.out.println(getShortestFragment("aZABcabbCa"));
    }

    static String getShortestFragment(String str){
        for(int k=1;k<=str.length();k++){
            for(int i=0;i<str.length()-k+1;i++){
                Set<Character> lowerSet = new HashSet<>();
                Set<Character> upperSet = new HashSet<>();
                String temp = str.substring(i,i+k);
                char[] tempCharArr = temp.toCharArray();
                for(char ch : tempCharArr){
                if(Character.isLowerCase(ch))
                    lowerSet.add(ch);
                else
                    upperSet.add(ch);
                }
                if(containsAllElements(lowerSet, upperSet) && containsAllElements(upperSet, lowerSet)){
                    return temp;
                }
            }            
        }
        return "-1";   
    }

    static boolean containsAllElements(Set<Character> first, Set<Character> second){
        Set<Character> lower1 = new HashSet<>();
        Set<Character> lower2 = new HashSet<>();
        first.forEach((e) -> {
            lower1.add(Character.toLowerCase(e));
        });
       second.forEach((e) -> {
            lower2.add(Character.toLowerCase(e));
        });
        return lower1.containsAll(lower2);
    }
}

Regards

Thanks for quick replay

I will try it

Sure, You are welcome.