How to count the number of equal values in two arrays?

Question

for Example:

int[] a = [0,1,2,3,4,5];
int[] b = [3,4,5,6,7,8];

count = 3;

The arrays do not have to be consecutive numbers. How would I get the number of values that equal between these arrays?

edit: So I've attempted the following:

List<int[]> w = Arrays.asList(winning);
List<int[]> s = Arrays.asList(F1Select);            
w.retainAll(s);
int equalNums = w.size();

But I'm getting the following error for the retainAll line:

Exception in thread "AWT-EventQueue-0" java.lang.UnsupportedOperationException
    at java.util.AbstractList.remove(Unknown Source)
    at java.util.AbstractList$Itr.remove(Unknown Source)
    at java.util.AbstractCollection.retainAll(Unknown Source)

What have you tried? The basic logic is to keep track of the ones you've already visited. — Sotirios Delimanolis, Mar 27 '13 at 18:31
Use a nested loop and are the values unique in one array or do you have duplicates? — Leonard Brünings, Mar 27 '13 at 18:32
Another important aspect: are the values in an array unique ? — Radu Murzea, Mar 27 '13 at 18:32
@SotiriosDelimanolis - What do you mean `The basic logic is to keep track of the ones you've already visited. `? — user93353, Mar 27 '13 at 18:34
@user93353 Having visited the first array, you know you've seen the numbers {0,1,2,3,4,5}. Maybe you put them in a set. When you go through the second array, you just check if the number is in that set. — Sotirios Delimanolis, Mar 27 '13 at 18:40
Why should we solve this question if the user has not taken any initiative to solve it..?? Show some effort from your behalf.. — Vishal K, Mar 27 '13 at 18:41

Ben McCormick · Answer 1 · 2013-03-27T18:42:47.700

4

You can just convert to lists, and find the intersection using retainAll.

List<Integer> aList =  Arrays.asList(a);
List<Integer> bList =  Arrays.asList(b);
aList.retainAll(bList);
return aList.size();

aList will then only contain items that are also in bList, and the size of aList lets you know the count.

If you want only unique values you could convert the arrays to a Set and do the same thing.

edited Mar 27 '13 at 18:42

answered Mar 27 '13 at 18:35

Ben McCormick

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I'm using Eclipse and it's telling me to us List instead of List. – Gordon Li Mar 27 '13 at 18:43

score 1 · Answer 2 · edited May 23 '17 at 12:21

1

Try this :

 Integer[] a = new Integer[]{0, 1, 2, 3, 4, 5};
 Integer[] b = new Integer[]{3, 4, 5, 6, 7, 8};

  List<Integer> list1 = Arrays.asList(a);
  Set<Integer> commonSet = new TreeSet<Integer>();
     for (Integer i : b) {
         if (list1.contains(i)) {
           commonSet.add(i);
            }
        }

        System.out.println(commonSet.size());

edited May 23 '17 at 12:21

Community

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answered Mar 27 '13 at 18:43

Alya'a Gamal

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score 0 · Answer 3 · answered Mar 27 '13 at 18:36

If your are allowed to have O(m + n) additional space used during computation, you can keep a HashMap for each array. The key is each array element, the value is how many times it occurs. Once you've computed how often each number occurs, then you've simplified the problem to comparing the two maps.

Whenever you have a key present in both maps, you have a number that is present in both arrays. The values can let you decide how many times the number is present in both arrays.

How to count the number of equal values in two arrays?

3 Answers3

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