Multiples of an element in a Java Array

Question

I have a pretty generic question about arrays. It is for a dice rolling game I have to make for homework.

I have an array of length 7 that contains ints that are the result of my various dice rolling methods and I need to find out if the array contains multiple instances of an element, what that element is, and how many times it occurs. Being new to Java, I don't even know where to look for a method that does this sort of thing. Does anyone know where I can find one, or better yet have tips on how to write one myself?

One of the hardest parts for me conceptually is that I don't know how many results I will have, as there could be up to two pairs and a trio. IE int[] roll = { 3, 3, 4, 4, 5, 5, 5 }. I think the best way to get around it would be to use a loop until my multiple finding method fails, each time removing the matching elements.

Thanks a lot for any help, this is due at midnight!

Are you rolling a single 6-sided die or a pair of 6-sided dice? (Will the results be 1-6 or 2-12?) — nhgrif
– nhgrif, Commented Oct 30, 2013 at 0:47
Use a Map instead, with key being the dice, and value being number of occurrences. — Sadek Noureddine
– Sadek Noureddine, Commented Oct 30, 2013 at 0:54

arshajii · Accepted Answer · 2013-10-30 00:52:22Z

1

The usual approach would be to use a Map to accumulate frequencies:

int[] roll = { 3, 3, 4, 4, 5, 5, 5 };

Map<Integer, Integer> counts = new HashMap<>(roll.length);

for (int a : roll)
    counts.put(a, counts.containsKey(a) ? counts.get(a) + 1 : 1);

for (Entry<?, Integer> e : counts.entrySet())
    if (e.getValue() > 1)
        System.out.println(e.getKey() + "\t" + e.getValue());

3   2
4   2
5   3

The first column is dice rolls and the second column is frequencies.

answered Oct 30, 2013 at 0:52

arshajii

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2 Comments

El Capitan Over a year ago

This looks great, Thank you. Is there a way to access the numbers in the map? As in, assign the output to variables to be used later?

arshajii Over a year ago

@user2812871 You can use counts.get(n) to get the frequency of number n. Also, if this answer helped you, you could consider accepting it.

tbodt · Accepted Answer · 2013-10-30 00:52:00Z

0

You shouldn't be using an array for this. The best thing to use is a Map<Integer, Integer>. The keys are the numbers, the values are the number of occurrences. Here's example code to keep track of the numbers like this:

Map<Integer, Integer> numbers = new Map<Integer, Integer>();
public void addNumber(int n) {
    if (numbers.get(n) == null)
        numbers.set(n, 0);
    int count = numbers.get(n);
    numbers.set(n, count + 1);
}

public int countForNumber(int n) {
    if (numbers.get(n) == null)
        return 0;
    return numbers.get(n);
}

answered Oct 30, 2013 at 0:52

tbodt

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Comments

nhgrif · Accepted Answer · 2013-10-30 01:01:21Z

0

int[7] rollCounts = {0};
for(int i=0; i<roll.length; ++i)
    rollCounts[roll[i]]++;

Using this method, rollCounts[1] will contain the count of occurrences of 1 in roll[], and rollCounts[2] will contain the count of occurrences of 2 in roll[] etc. (rollCounts[0] should always be 0 assuming roll[i] never contains a 0, which it shouldn't).

Then you can just use some logic to check rollCounts[i] in a different for loop to do whatever you need to do based on the number of occurrences.

And honestly, you could just store your dice rolls in this array, unless it's important to keep track of the order the rolls occurred in.

edited Oct 30, 2013 at 1:01

answered Oct 30, 2013 at 0:51

nhgrif

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Comments

Rock · Accepted Answer · 2013-10-30 01:38:22Z

0

int[] arr = { 2, 2, 4, 4, 4, 6, 6, 6, 6 };  // Can have any length.
Map<Integer, Integer> elemCount = new HashMap<Integer, Integer>();
Integer count = 0;
for ( int element : arr )
{
    count = elemCount.get ( (Integer) element );
    if ( count == null )
    {
        count = 1;
    }
    else 
    {
        count++;
    }
    elemCount.put ( (Integer) element , (Integer) count );
}
System.out.println ( elemCount.toString() );

answered Oct 30, 2013 at 1:38

Rock

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