Converting a class interface into a class template - c++

Question

I'm working through some old exam papers without the memos. I just want to make sure I am understanding this correctly.
Provide an interface of the template class
Dictionary <Tkey, TValue>
This is the scenario provided
class Dictionary {
public:
Dictionary();
void Add(int key, const string &value);
string Find (int key) const;
private:
vector<int> Keys;
vector <string> Values;
};
This is the solution I have written down
class Dictionary {
public:
Dictionary();
void Add(TKey key, const TValue &value);
TValue Find (TKey key) const;
private:
vector <Dictionary> Keys;
vector <Dictionary> Values;
};
This appears correct to me. I haven't compiled a driver program for this as I just want to make sure I am understand this correctly given the template class.
I think just the last 2 lines where vector is concerned is what I want to make sure I've written correctly.
Thanks for taking the time.

Answer 1

You should simply follow your instructions:
template<typename Tkey, typename TValue> // <<<<<<<<
class Dictionary {
public:
Dictionary();
void Add(TKey key, const TValue &value);
TValue Find (TKey key) const;
private:
vector <TKey> Keys; // <<<<<<<<
vector <TValue> Values; // <<<<<<<
};
Or even better (since it's going to be hard to associate those vector members appropriately):
template<typename Tkey, typename TValue> // <<<<<<<<
class Dictionary {
public:
Dictionary();
void Add(TKey key, const TValue &value);
TValue Find (TKey key) const;
private:
vector <std::pair<TKey,TValue>> dict; // <<<<<<<
};

Answer 2

This conversion is incomplete, and slightly incorrect.
To make it complete, make sure the class is actually a class template, i.e. there is
template <typename TKey, typename TValue>
class Dictionary {
...
};
The correction is to make the two vectors take keys and values. Currently, both vectors are set to store Dictionary elements, which is not what you need: the first vector should hold TKey elements, while the second one should hold TValues. You would discover this shortcoming as soon as you'd start implementing your Dictionary<TKey,TValue>::Find method.