Comparing elements of object arrays

Question

Lets say I have a class called "Sequence". This class has an instance variable private int[] tab. There are also some methods to create this tab like those:

public Sequence() {
    this.tab = fillAnArray(drawsNumber(5, 20));
}

private int drawsNumber(int min, int max) {
    if (min > max) {
        throw new IllegalArgumentException("Wrong range");
    }
    return new Random().nextInt(max - min + 1) + min;
}

public int[] fillAnArray(int size) {
    int[] arr = new int[size];
    arr[0] = 1;
    for (int i = 1; i < arr.length; i++) {
        arr[i] = drawsNumber(arr[i - 1], arr[i - 1] + 10);
    }
    return arr;
}

Now, I would like to create the method that accepts two Sequence objects as arguments and returns number of the same tab elements. So, I created method like this:

public int howManyCommonElements(Sequence c1, Sequence c2) {
    int i = 0, j = 0;
    int counter = 0;
    while (i < c1.tab.length && j < c2.tab.length) {
        if (c1.tab[i] == c2.tab[j]) {
            ++counter;
        } else if (c1.tab[i] < c2.tab[j]) {
            ++i;
        } else {
            ++j;
        }
    }
    return counter;
}

How do I compare single elements like c1.tab[i], c2.tab[j]?

What do you mean you don't know how to compare? What is wrong with c1.tab[i] == c2.tab[j]? BTW it is bug in your code, if tabs have common element your loop will never finish. — talex
– talex, Commented Sep 28, 2020 at 7:40
Ouch, i just tried to use something different than for loop. Youre right, i do not know how i could not see the problem. — viverte
– viverte, Commented Sep 28, 2020 at 7:50
Does this answer your question? finding common elements in two integer arrays java — Abra
– Abra, Commented Sep 28, 2020 at 7:52

LucaBonadia · Accepted Answer · 2020-09-28 07:40:39Z

0

You have to compare every element of the 1st array with every element of the 2nd one. To do that you have to use 2 nested for cycles:

for (int i = 0; i < c1.tab.length; i++)
    for (int j = 0; j < c2.tab.length; j++)
        if (c1.tab[i] == c2.tab[j])
            ++counter;

answered Sep 28, 2020 at 7:40

LucaBonadia

913 silver badges14 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

talex Over a year ago

No, elements are ordered, you don't need to make it O(n*m) out of existing O(n+m).

SSC · Accepted Answer · 2020-09-28 08:30:04Z

0

If you have ordered elements (ASC), you can do like;

    int i = 0, j = 0;
    int counter = 0;
    while (i < arr1.length && j < arr2.length) {
        if (arr1[i] == arr2[j]) {
            ++i;
            ++j;
            ++counter;
        } else if (arr1[i] < arr2[j]) {
            ++i;
        } else {
            ++j;
        }
    }

edited Sep 28, 2020 at 8:30

answered Sep 28, 2020 at 7:43

SSC

3,0183 gold badges29 silver badges43 bronze badges

1 Comment

dbl Over a year ago

Dude, the elements are already ordered asc... Idk if you noticed but the OP resolved the issue from the comments, even before the answers added here.

Collectives™ on Stack Overflow

Comparing elements of object arrays

2 Answers 2

1 Comment

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related