[FIXED] JAVA : how to reduce code size, when using lambda to process array's items

Issue

I must sort each item in the array, in alphabatic order

in : [bcdef, dbaqc, abcde, omadd, bbbbb] out : [bcdef, abcdq, abcde, addmo, bbbbb]

I wrote the code below but i feel it verbose(long).

Could you please tell me an other way, with shorter code ?

Thanks.

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;

import static java.lang.String.copyValueOf;

public class Main {

    public static void main(String[] args) {
        String[] stringsArray = { "bcdef", "dbaqc", "abcde", "omadd", "bbbbb"};

        System.out.println(Arrays.toString(stringsArray));

        List<String> list = Arrays.stream(stringsArray).map((String s)->{
            char[] charArray = s.toCharArray();
            Arrays.sort(charArray);
            return copyValueOf(charArray);
        }).collect(Collectors.toList());
        ArrayList<String> arrayList = new ArrayList<String>(list);
        stringsArray = Arrays.copyOf(arrayList.toArray(),arrayList.size(),String[].class);

        System.out.println(Arrays.toString(stringsArray));
    }

}

Solution

It looks like you want to sort characters within each element, not the whole array.

You could simplify by having separate map() (and peek() since Arrays.sort() method is not returning a value) steps. You could also skip redundant List by collecting directly to a String array.

String[] sorted = Arrays.stream(stringsArray)
    .map(String::toCharArray)
    .peek(Arrays::sort)
    .map(String::new)
    .toArray(String[]::new);

Answered By - Karol Dowbecki
Answer Checked By - Katrina (JavaFixing Volunteer)