CODEWITHSUNDEEP
javasortingcomparator
Herschelle
Herschelle

Reputation: 27

Sort character array using java comparator

Sort a character array lexicographically with an additional condition that all c's comes before all b's. This could be done manually but I want to code it via inbuilt sorting using comparators. The code I wrote -

static class Sort implements Comparator<Character> {
    @Override
    public int compare(Character x, Character y) {
        if(x == 'c' && y == 'b') {
            return y - x;
        }
        return x - y;
    }
}

public static void main(String[] args) {
    String s = "abracadabra";
    int n = s.length();
    Character[] res = new Character[n];
    for (int i = 0; i < n; i++) {
        res[i] = s.charAt(i);
    }
    Arrays.sort(res, new Sort());
    System.out.println(Arrays.toString(res));
}

Gives output: [a, a, a, a, a, b, c, b, d, r, r]. The c only comes before one of the b's. If I change the comparator to the following then it gives the correct output.

public int compare(Character x, Character y) {
        if(x == 'c' && y == 'b' || x == 'b' && y == 'c') {
            return y - x;
        }
        return x - y;
    }

My question is why in both cases returning "y-x" gives the correct answer? If in one case returning "y-x" was correct then in the other case shouldn't returning "x-y" would have been correct? Instead of "y-x", returning a -1 or +1 doesn't work either. Pl. explain what is happening internally. Thanks!

Upvotes: 1

Views: 1146

Answers (3)

tquadrat
tquadrat

Reputation: 4034

See the Javadoc for java.util.Comparator. It says for the method compare():

Compares its two arguments for order. Returns a negative integer, zero, or a positive integer as the first argument is less than, equal to, or greater than the second.

For your case, it means that compare( 'b', 'c' ) should return something greater than zero, and for compare( 'c', 'b' ) something less than zero, while the natural order would be the opposite.

But your first implementation (that wrong one) covered only the case compare( 'c', 'b' ), but not compare( 'b', 'c' ), so that the method would return the default value then.

Your second and correctly working implementation fixed that.

Upvotes: 0

Agzam
Agzam

Reputation: 138

You can add System.out.println() to understand how it work:

Code

import java.util.Arrays;
import java.util.Comparator;

public class Main {

    public static void main(String[] args) {
        String s = "abracadabra";
        int n = s.length();
        
        System.out.println("X Y\tX Y\t[if] VALUE");
        System.out.println();
        
        Character[] res = new Character[n];
        for (int i = 0; i < n; i++) {
            res[i] = s.charAt(i);
        }
        
        int min = 'a';
        
        Arrays.sort(res, new Comparator<Character>() {

            @Override
            public int compare(Character x, Character y) {
                System.out.print(y + " " + x + "\t" + (x-min) + " " + (y-min) + "\t");
                if(x == 'c' && y == 'b') {
                    System.out.println("true " + (y - x));
                    return y - x;
                }
                System.out.println("     " + (x - y));
                return x - y;
            }
        });
        System.out.println("#################################\nResult:\n" + Arrays.toString(res));
    }
}

Console:

X Y X Y   [if] VALUE

a b 1 0        1
b r 17 1       16
r a 0 17       -17
b a 0 1        -1
a a 0 0        0
b c 2 1   true -1
a c 2 0        2
c a 0 2        -2
a a 0 0        0
c d 3 2        1
r d 3 17       -14
b d 3 1        2
c a 0 2        -2
a a 0 0        0
a a 0 0        0
c b 1 2        -1
a b 1 0        1
a b 1 0        1
b r 17 1       16
d r 17 3       14
r r 17 17      0
c a 0 2        -2
a a 0 0        0
b a 0 1        -1
a a 0 0        0
#################################
Result:
[a, a, a, a, a, b, c, b, d, r, r]

Upvotes: 1

Shawn
Shawn

Reputation: 52344

The problem is that if passing 'c' and 'b' makes the comparison return a negative value, then passing 'b' and 'c' should return a positive one (And your first version returns a negative number in that case instead). If the function doesn't return consistent results no matter the order of arguments, the sort algorithm is going to produce a garbage order.

Consider this version:

    public int compare(Character x, Character y) {
        if (x == 'c' && y == 'b') {
            return -1;
        } else if (x == 'b' && y == 'c') {
            return 1;
        } else {
            return x.compareTo(y);
        }
    }

Upvotes: 4

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