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!

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;
}

python 3 solution using hash map, pass all cases

def matchingStrings(strings, queries):
        d = {}
        for s in queries:
                d[s] = 0
        for ss in strings:
                if ss in d:
                        d[ss]+=1
        ans = [d[num] for num in queries]
        return ans

I feel this is the simplest approach to this problem.

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        String[] a = new String[N];
        for(int i=0; i<N; i++){
            a[i] = sc.next();
        }
        
        int Q = sc.nextInt();
        int[] result = new int[Q];
        int i=0;
        
        while(sc.hasNext()){
            String query = sc.next();
            int count = 0;
            for(int j=0; j<N; j++){
                if (a[j].equals(query)){
                    count++;
                }
            }
            result[i++] = count;
        }
        
        for(i=0; i<Q; i++){
            System.out.println(result[i]);
        }
    }
}