How can I find the smallest covering prefix of an array in Java?

Question

Find the first covering prefix of a given array.

A non-empty zero-indexed array A consisting of N integers is given. The first covering prefix of array A is the smallest integer P such that and such that every value that occurs in array A also occurs in sequence.

For example, the first covering prefix of array A with A[0]=2, A[1]=2, A[2]=1, A[3]=0, A[4]=1 is 3, because sequence A[0], A[1], A[2], A[3] equal to 2, 2, 1, 0 contains all values that occur in array A.

My solution is

int ps ( int[] A ) 
{
    int largestvalue=0;
    int index=0;   

    for(each element in Array){
        if(A[i]>largestvalue)
        {
            largestvalue=A[i];
            index=i;
        }
    }

    for(each element in Array)
    {
        if(A[i]==index)
            index=i; 
    }   
    return index;
}

But this only works for this input, this is not a generalized solution.

Answer 1

Got 100% with the below.

public int ps (int[] a)
    {
        var length = a.Length;
        var temp = new HashSet<int>();
        var result = 0;

        for (int i=0; i<length; i++)
        {
            if (!temp.Contains(a[i]))
            {
                temp.Add(a[i]);
                result = i;
            }
        }
        return result;
    }

Answer 2

I would do this

int coveringPrefixIndex(final int[] arr) {
    Map<Integer,Integer> indexes = new HashMap<Integer,Integer>();
    // start from the back
    for(int i = arr.length - 1; i >= 0; i--) {
        indexes.put(arr[i],i);
    }
    // now find the highest value in the map
    int highestIndex = 0;
    for(Integer i : indexes.values()) {
        if(highestIndex < i.intValue()) highestIndex = i.intValue();
    }
    return highestIndex;
}