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 <= 106
• At most 104 calls will be made to put, get, and remove.

Problem Link

Design HashMap - LeetCode

Code

class MyHashMap {
    
    vector<vector<int>> myHash;
      // Create a vector containing n 
    //vectors of size m.  
    //vector<vector<int> > vec( n , vector<int> (m, 0));
    
public:
    /** Initialize your data structure here. */
    MyHashMap() {
         // resize
        myHash.resize(100000, vector<int>(10, -1));
        // initialize
       // vector<int> temp(10,-1);
       //  for(int i = 0; i < 100000; i++ )
       //      myHash.push_back(temp);
    }
    
    /** value will always be non-negative. */
    void put(int key, int value) {
        myHash[key / 10][key % 10] = value;
    }
    
    /** Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key */
    int get(int key) {
        return myHash[key / 10][key % 10];
    }
    
    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    void remove(int key) {
        myHash[key / 10][key % 10] = -1;
    }
};