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Design HashMap

Problem Description​

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(int key) removes the key and its corresponding value if the map contains the mapping for the key.

Examples​

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 for Design HashMap​

Approach​

Array-Based Implementation​

  • Array Initialization: Use an array to store the values with a size large enough to accommodate the given constraints.
  • Direct Indexing: Use the key as an index to store the value in the array.
  • Handling Removal: Use a special value (like -1) to denote the removal of a key.

Implementation:

class MyHashMap:

    def __init__(self):
        self.size = 1000001
        self.hashmap = [-1] * self.size

    def put(self, key: int, value: int) -> None:
        self.hashmap[key] = value

    def get(self, key: int) -> int:
        return self.hashmap[key]

    def remove(self, key: int) -> None:
        self.hashmap[key] = -1

# Example usage
myHashMap = MyHashMap()
myHashMap.put(1, 1)
myHashMap.put(2, 2)
print(myHashMap.get(1))    # returns 1
print(myHashMap.get(3))    # returns -1 (not found)
myHashMap.put(2, 1)        # update the existing value
print(myHashMap.get(2))    # returns 1
myHashMap.remove(2)        # remove the mapping for 2
print(myHashMap.get(2))    # returns -1 (not found)

Complexity:

  • Time Complexity: O(1) for put, get, and remove operations.
  • Space Complexity: O(n), where n is the range of keys (1,000,001).

Corner Cases:

  • Handling updates: Ensure that the value is updated if the key already exists.
  • Handling removals: Ensure that the key is marked as removed.

Code in Different Languages​

Written by @vansh-codes
var MyHashMap = function() {
    this.size = 1000001;
    this.map = new Array(this.size).fill(-1);
};

MyHashMap.prototype.put = function(key, value) {
    this.map[key] = value;
};

MyHashMap.prototype.get = function(key) {
    return this.map[key];
};

MyHashMap.prototype.remove = function(key) {
    this.map[key] = -1;
};

References​