Find odd numbers between a closed range

Question

I am doing a code challenge where given two integers l and r, I have to print all the odd numbers between i and r (i and r inclusive). The function must return an array of integers denoting the odd numbers between l and r.

This is what I have so far

 static List<Integer> oddNumbers(int l, int r) {
    List<Integer> list1 = new ArrayList<>();
    if ((r > l) && (l >= 1) && (r <= 100000)) {
        for (int i = tmp; tmp < r; i++) {
            if (l % 2 == 0)
               l = l + 1;              
            list1.add(l);
            l = l + 2;
        }
    }
    return list1;
}

However, I received several errors like this

Compiler Message
Wrong Answer
Your Output (stdout)
Output hidden

Any ideas for this? on the QA of the challenge site, it seems related to Corner case problems.

these seem like custom messages. where are they in your code? I would recommend not to duplicate code. yes, you have to do a even check before adding, but you don't need to duplicate all the following code — Stultuske, Mar 12 '19 at 08:17
looks like you are excluding r. Besides, should you return an array or a List? It's unclear. — Eran, Mar 12 '19 at 08:19
@Eran quite possibly that's the reason for the Compiler Message - error — Stultuske, Mar 12 '19 at 08:20

Karim · Answer 1 · 2019-03-12T10:18:54.270

4

I would suggest the following code :

public static List<Integer> getOdd(int l, int r) {
    List<Integer> odd = new ArrayList<Integer>();
    if (l % 2 == 0)
        l++;
    while(l <= r) {
        odd.add(l);
        l += 2;
    }
    return odd;
}

edited Mar 12 '19 at 10:18

answered Mar 12 '19 at 08:27

Karim

187
10

1

seriously? why? first of all, it has to return an array, not a list, secondly, why would you want to compare all values to check if they are odd/even, if you can just get the first odd one, and add 2 each iteration, instead of re-checking every time? – Stultuske Mar 12 '19 at 09:51
I saw in @Yang 's example that he was returning a list. And yeah you're right, it's better to get the first odd number and add 2. – Karim Mar 12 '19 at 10:13
Indeed, he returns a list. He also receives a compiler error message, which might very well be caused by returning a list. – Stultuske Mar 12 '19 at 10:15
I did edit my answer to avoid re-checking every time, but for the List I really don't know because he didn't specify anything. Anyway thanks for your answers ! – Karim Mar 12 '19 at 10:20
as his requirement description states: "The function must return an array of integers" – Stultuske Mar 12 '19 at 10:21

score 1 · Accepted Answer · answered Mar 12 '19 at 08:36

1

As the comments have mentioned, you're excluding r in your loop and are perhaps returning the wrong data structure?

Moreover, your code is doing way more than it needs to

you only need one loop
you don't need tmp at all - just modify l
you don't need all the if/else blocks

I suggest the following:

static int[] oddNumbers(int l, int r) {
    List<Integer> list1 = new ArrayList<>();
    if (l % 2 == 0) l++;
    for(; l <= r; l += 2) {
        list1.add(l);
    }
    return list1.stream().mapToInt(i->i).toArray(); // https://stackoverflow.com/a/23945015/2554605
}

You can add back the if ((r >= l) && (l >= 1) && (r <= 100000)) if it's needed based on challenge requirements, but I'm not that it's necessary (notice I have r >= l instead of just r > l to cover the case where r == l).

As well, if you really are allowed to return List<Integer> instead of int[], then just return list1; instead as you had it.

answered Mar 12 '19 at 08:36

Michael Yaworski

13,410
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just a thought. so far as I can tell, the limits (inclusive) are l and r, but unless the requirement on that page is different compared to the one posted here, it's not stated which of the two is the upper, and which the lower limit. – Stultuske Mar 12 '19 at 10:16
@Stultuske I figure the `l` and `r` are self-explanatory due to their names. I suppose if it were generic `a` and `b` though, we could swap the values if `a > b` and keep our logic the same. – Michael Yaworski Mar 12 '19 at 12:09
how are i and r self-explanatory? might be because I myself am not a native English speaker, but still .. – Stultuske Mar 12 '19 at 12:13
@Stultuske I believe it's `l`, not `i`. I know OP said `i` but it really doesn't make sense because he starts with `l`. And `l` is short for `left`; `r` is short for `right`. So `r` is upper bound is `l` is lower bound. – Michael Yaworski Mar 12 '19 at 12:15

score 0 · Answer 3 · answered Mar 12 '19 at 08:33

0

this version should be working

 static List<int> oddNumbers(int l, int r)
    {
        List<int> list1 = new List<int>();
        if ((r > l) && (l >= 1) && (r <= 100000))
        {
            int tmp = l % 2 == 0 ? l + 1 : l;            
            for (int i = tmp; i < r; i += 2)
            {
                list1.Add(i);
            }

        }
        return list1;
    }

answered Mar 12 '19 at 08:33

Oleg Bondarenko

1,694
1
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No, it shouldn't. for starters: it returns a List – Stultuske Mar 12 '19 at 09:50
adds ToArray() to result, but it is algorithm question. – Oleg Bondarenko Mar 12 '19 at 10:17
No. the question is: "where do my errors come from". just the title is limited to that – Stultuske Mar 12 '19 at 10:18
Array in this case is overusing memory it might regard as error. Or only logical issues should be noted? – Oleg Bondarenko Mar 12 '19 at 10:24
depends on the requirements. If the requirements are: "return an array", chances are the result will be ran against a number of asserts to check whether it is an array, and whether result.length is the value they expect. – Stultuske Mar 12 '19 at 10:35
I got your point , thanks. But I hope it is site for sharing knowledge but not for red tape. – Oleg Bondarenko Mar 12 '19 at 12:46

score 0 · Answer 4 · edited Mar 24 '21 at 14:12

0

Instead of dividing by 2 and checking if the remainder is 0 or 1, you can bitwise and with 1(&1) and it will look like this if ((number&1)==1) then the number is odd, it reduce complexity.

edited Mar 24 '21 at 14:12

bguiz

27,371
47
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answered Mar 23 '21 at 02:14

INDERJEEt

1

Find odd numbers between a closed range

4 Answers4