Problem statement

Design a HashMap without using any built-in hash table libraries.

Implement the MyHashMap class:

• MyHashMap() initializes the object with an empty map.
• void put(int key, int value) inserts a (key, value) pair into the HashMap. If the key already exists in the map, update the corresponding value.
• int get(int key) returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
• void remove(key) removes the key and its corresponding value if the map contains the mapping for the key.

Example 1

Input
["MyHashMap", "put", "put", "get", "get", "put", "get", "remove", "get"]
[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
Output
[null, null, null, 1, -1, null, 1, null, -1]

Explanation
MyHashMap myHashMap = new MyHashMap();
myHashMap.put(1, 1); // The map is now [[1,1]]
myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
myHashMap.get(1);    // return 1, The map is now [[1,1], [2,2]]
myHashMap.get(3);    // return -1 (i.e., not found), The map is now [[1,1], [2,2]]
myHashMap.put(2, 1); // The map is now [[1,1], [2,1]] (i.e., update the existing value)
myHashMap.get(2);    // return 1, The map is now [[1,1], [2,1]]
myHashMap.remove(2); // remove the mapping for 2, The map is now [[1,1]]
myHashMap.get(2);    // return -1 (i.e., not found), The map is now [[1,1]]

Constraints

• 0 <= key, value <= 10^6.
• At most 10^4 calls will be made to put, get, and remove.

Solution 1: Store all keys' values

Code

#include <iostream>
#include <vector>
using namespace std;
class MyHashMap {
    vector<int> _v;
public:
    MyHashMap() : _v(1000001, -1) {

    }
    void put(int key, int value) {
        _v[key] = value;
    }
    int get(int key) {
        return _v[key];
    }
    void remove(int key) {
        _v[key] = -1;
    }
};
int main() {
    MyHashMap m;
    m.put(1, 1);
    m.put(2, 2);
    cout << m.get(1) << endl;
    cout << m.get(3) << endl;
    m.put(2, 1);
    cout << m.get(2) << endl;
    m.remove(2);
    cout << m.get(2) << endl;
}

Output:
1
-1
1
-1

Complexity

• Runtime: O(1).
• Extra space: KEY_MAX, which is 10^6 in this problem.

Solution 2: Store only real keys

Code

#include <iostream>
#include <vector>
using namespace std;
class MyHashMap {
    vector<pair<int, int> > _v;
public:
    MyHashMap() {

    }
    void put(int key, int value) {
        for (auto& p : _v) {
            if (p.first == key) {
                p.second = value;
                return;
            }
        }
        _v.push_back(make_pair(key, value));
    }
    int get(int key) {
        for (auto& p : _v) {
            if (p.first == key) {
                return p.second;
            }
        }
        return -1;
    }
    void remove(int key) {
        auto it = _v.begin();
        while (it != _v.end()) {
            if (it->first == key) {
                _v.erase(it);
                return;
            } else {
                it++;
            }
        }
    }
};
int main() {
    MyHashMap m;
    m.put(1, 1);
    m.put(2, 2);
    cout << m.get(1) << endl;
    cout << m.get(3) << endl;
    m.put(2, 1);
    cout << m.get(2) << endl;
    m.remove(2);
    cout << m.get(2) << endl;
}

Output:
1
-1
1
-1

Complexity

• Runtime: O(N), where N is the number of keys.
• Extra space: O(2N).