Google Foobar challenge Minion Work Assignment Java test cases not passing

Question

I have tried solving this problem using Java and for some reason only 2 out of 9 test cases pass but locally all my test cases pass. I am 99% positive that there is some issue with Google Foobar Java test cases for this challenge. Did someone encounter this and if yes, what did you do to solve it?

Question was...

Write a function called solution(data, n) that takes in a list of 
less than 100 integers and a number n, and returns that same list 
but with all of the numbers that occur more than n times removed 
entirely. 

The returned list should retain the same ordering as the original 
list - you don't want to mix up those carefully-planned shift 
rotations! For instance, if data was [5, 10, 15, 10, 7] and n was 1,
solution(data, n) would return the list [5, 15, 7] because 10 occurs 
twice, and thus was removed from the list entirely.

-- Java cases --
Input:
Solution.solution({1, 2, 3}, 0)
Output:
    

Input:
Solution.solution({1, 2, 2, 3, 3, 3, 4, 5, 5}, 1)
Output:
    1,4

There are 6 more test cases that are hidden.

Below is the solution I created.

public class MinionShift {
    public static int[] solution(int[] data, int n) {
        if(n<1)
            return new int[0];

        if(data.length < 1)
            return data;

        Map<Integer, Integer> map = new HashMap<>();

        for(int d: data) {
                map.put(d, map.getOrDefault(d, 0) + 1);
        }

        return Arrays.stream(data).filter(c->map.containsKey(c) && !(map.get(c)>n)).toArray();
    }
}

Test cases that I have tried...

[{1, 2, 3}, 0]
[{1, 2, 2, 3, 3, 3, 4, 5, 5}, 1]
[{1, 2, 2, 3, 3, 3, 4, 5, 5}, 10]
[{1, 2, 2, 3, 3, 3, 4, 5, 5}, -1]
[{}, 5]
[{1, 1, 1, 1, 1}, 5]
[{101, 102, 103, 104, 105}, 5]

Edit... I tried a Java stream based solution as follows but unfortunately the challenge went away as I submitted a Python solution. But I am posting it here anyways.

public class MinionShift {
    public static int[] solution(int[] data, int n) {
        if(n<1)
            return new int[0];

        if(data.length < 1)
            return data;

        return Arrays.stream(data).filter(d->Arrays.stream(data).filter(i->i==d).count()<=n).toArray();
    }
}

Aside: `map.merge(d, 1, Integer::sum);` is a neater way to build the map of counts. — Andy Turner, Aug 13 '21 at 16:59
(and `map.getOrDefault(c, 0)>=n` is a neater filter function at the end. Or just `get`, `c` will always be in the map). — Andy Turner, Aug 13 '21 at 17:04

score 0 · Answer 1 · answered Aug 13 '21 at 17:20

This is how I got it to work, you have to work backwards with the array. You need to create another array to check for duplicates to remember them.

public static int[] solution(int[] data, int n){
    if(n < 1){
        return new int[0];
    }

    if(data.length < 1){
        return data;
    }

    List<Integer> a = Arrays.stream(data).boxed().collect(Collectors.toList());

    for(int i = a.size()-1; i > -1; i--){
        ArrayList<Integer> t = new ArrayList<>();

        for(int j = 0; j < a.size(); j++){
            if(a.get(j) == a.get(i)){
                t.add(j);
            }
        }


        if(t.size() > n){
            for(int j = t.size()-1; j > -1; j--){
                a.remove((int) t.get(j));
            }

            i -= t.size()-1;
        }
    }

    data = new int[a.size()];
    int c = 0;
    for(int d : a){
        data[c] = d;
        c++;
    }

    return data;
}

Google Foobar challenge Minion Work Assignment Java test cases not passing

1 Answers1