Counting unique values in an array (Java)

Question

It's been asked before, I know, but I really don't know why my code isn't working. I'm trying to figure out how many unique values are in an undefined array. (i.e. {0, 0, 1, 1, 1, 5, 5, 6, 6, 7} should return a value of 5. This is what I have so far:

public static int numUnique(double[] list) {
    int counter = 0;
    
    for(int i = 0; i < list.length; i++) {
        for(int j = i + 1; j < list.length; j++) {
            if(list[i] != list[j])
                counter ++;
        }
    }
    
    
    return counter;
}

HashMap is really good for this if the value is already inside the HashMap than count it's value up otherwise create a new entry. Then check how many of the keys have value 1 — Aalexander, Jan 08 '21 at 22:28
@PetterFriberg, suggested answer refers to javascript not java — Nowhere Man, Jan 08 '21 at 22:44

score 7 · Accepted Answer · edited Mar 12 '21 at 15:33

7

Assuming that the array is sorted, you should only be checking if each element is not equal to the element immediately after it. The counter should also be initialized to 1 since if all elements are the same, the number of unique elements is 1. We also need to add a check for a null or empty array, for which the result is 0. (If it isn't sorted, you could use Arrays.sort to sort it, but that is not the fastest method.)

public static int numUnique(double[] list) {
    if(list == null || list.length == 0) return 0;
    int counter = 1;
    
   for(int i = 1; i < list.length; i++)
       if(list[i - 1] != list[i]) ++counter;
    
    
    return counter;
}

Alternate methods include using Stream#distinct or a Set, which do not rely on order^*.

System.out.println(java.util.Arrays.stream(array).distinct().count());

or

System.out.println(java.util.stream.IntStream.of(array).distinct().count());

^{* LinkedHashSet implementation provides you with predictable iteration order.}

edited Mar 12 '21 at 15:33

Arvind Kumar Avinash

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answered Jan 08 '21 at 22:27

Unmitigated

76,500
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why does this works? when you just check the one after it? – Aalexander Jan 08 '21 at 22:31
@Alex It is assumed that the array is sorted. – Unmitigated Jan 08 '21 at 22:32
I doubt that would work with `[0,1,0,1,0,1,0,1]`. The array is not said to be sorted. – Matthieu Jan 08 '21 at 22:32
1

@iota also OP doesn't mentioned that it is sorted – Aalexander Jan 08 '21 at 22:33
@Alex I've added an alternate method, in any case. – Unmitigated Jan 08 '21 at 22:35
@iota nice as a one liner. Have to try this – Aalexander Jan 08 '21 at 22:36
I should have mentioned, it is sorted. Thank you so much everyone, I've been scratching my head for hours, just started programming so the logic gets past me almost every time. – Ulises Pico Jan 08 '21 at 22:47

Nowhere Man · Answer 2 · 2021-01-08T22:39:26.537

2

If the input array is sorted as in your example, you do not need a nested loop:

public static int numUnique (double[] list) {
    if (null == list || 0 == list.length) {
        return 0;
    }
    int counter = 1;
    
    for (int i = 1; i < list.length; i++) {
        if (list[i] != list[i - 1]) {
            counter ++;
        }
    }
    
    
    return counter;
}

It is possible to use Stream API, functions distinct and count to achieve the same result (input array does not need to be sorted then):

public static int numUniqueStream (double ... list) {
    if (null == list) {
        return 0;
    }
    return (int) Arrays.stream(list).distinct().count();
}

edited Jan 08 '21 at 22:39

answered Jan 08 '21 at 22:31

Nowhere Man

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And if the array is not sorted...? – Matthieu Jan 08 '21 at 22:36
@Matthieu, it doesn't matter -- all of this is covered in iota's answer. I type a way too slow. – Nowhere Man Jan 08 '21 at 22:47

Counting unique values in an array (Java)

2 Answers2