Well, sadly, this problem isn't really about sparse arrays or whatever intention was. It's silly. If you read this comment, then i'm pretty sure, you have the same opinion. I propose to have some fun! Let's write shortest/cleanest code which solves this stuff in our favourite languages. Here's my for java8 :P

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        List<String> strings = IntStream.range(0, in.nextInt()).mapToObj(i -> in.next()).collect(Collectors.toList());
        IntStream.range(0, in.nextInt()).mapToObj(i -> in.next()).mapToLong(q -> strings.stream().filter(q::equals).count()).forEach(System.out::println);
    }

Come on guys =) why you think of it as the fewer lines or seconds you spent — you reach the better solution? This challenge is about fundamental computer science techniques, google about Sparse Arrays and make a decision why this challenge is named so (this is not without purpose). Believe me it's interesting and valuable =) HashMap is the right pill here and of course we would hardly find any programming language not giving HashMaps out of the box. But how HashMaps actually work under the hoods? Go through this!

This challenge is categorized as "Difficult"? I solved it in just a few minutes using a straight-ahead O(n^2) loop and it worked. Then I went back and re-did it with an associative array (hash map) in just a few more minutes and it worked as well. I don't think I learned anything about data structures or sparse arrays.

Here is my simple code for this:

int main() 
{
    int n,i,m,j;
    cin>>n;
    string s[n];
    for(i=0;i<n;i++)
    {
        cin>>s[i];
    }
    cin>>m;
    string t[m];
    for(i=0;i<m;i++)
    {
        cin>>t[i];
    }
    for(i=0;i<m;i++)
    {
        int flag=0;
        for(j=0;j<n;j++)
        {
            if(t[i]==s[j])
                flag++;
        }
        cout<<flag<<"\n";
    }
    return 0;
}

I used the map function in C++ with every string mapped to their occurence. Once I stored everything to the map I can just extract the number of occurences and print it out.