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Hash Map Without Using Built-in Hash Table Libraries

Certification: Intermediate Level Accuracy: 0% Submissions: 0 Points: 10

Write a Python program to design a hash map without using any built-in hash table libraries. Implement the MyHashMap class with put, get, and remove operations.

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: 
      

    • Step 1: MyHashMap hashMap = new MyHashMap() - Initialize an empty hash map.
      • 

    • Step 2: hashMap.put(1, 1) - Insert key-value pair (1, 1). Map becomes [[1,1]].
      • 

    • Step 3: hashMap.put(2, 2) - Insert key-value pair (2, 2). Map becomes [[1,1], [2,2]].
      • 

    • Step 4: hashMap.get(1) - Return value for key 1, which is 1.
      • 

    • Step 5: hashMap.get(3) - Key 3 doesn't exist, so return -1.
      • 

    • Step 6: hashMap.put(2, 1) - Update value for key 2 to 1. Map becomes [[1,1], [2,1]].
      • 

    • Step 7: hashMap.get(2) - Return value for key 2, which is now 1.
      • 

    • Step 8: hashMap.remove(2) - Remove key-value pair with key 2. Map becomes [[1,1]].
      • 

    • Step 9: hashMap.get(2) - Key 2 no longer exists, so return -1.
      • 

    

    •
Example 2
    

  • Input: ["MyHashMap", "put", "put", "put", "get", "remove", "get"]
[[], [10, 20], [30, 40], [10, 50], [10], [30], [30]]
    • 

  • Output: [null, null, null, null, 50, null, -1]
    • 

  • Explanation: 
      

    • Step 1: MyHashMap hashMap = new MyHashMap() - Initialize an empty hash map.
      • 

    • Step 2: hashMap.put(10, 20) - Insert key-value pair (10, 20). Map becomes [[10,20]].
      • 

    • Step 3: hashMap.put(30, 40) - Insert key-value pair (30, 40). Map becomes [[10,20], [30,40]].
      • 

    • Step 4: hashMap.put(10, 50) - Update value for key 10 to 50. Map becomes [[10,50], [30,40]].
      • 

    • Step 5: hashMap.get(10) - Return value for key 10, which is 50.
      • 

    • Step 6: hashMap.remove(30) - Remove key-value pair with key 30. Map becomes [[10,50]].
      • 

    • Step 7: hashMap.get(30) - Key 30 no longer exists, so return -1.
      • 

    

    •
Constraints
    

  • 0 ≤ key, value ≤ 10^6
    • 

  • At most 10^4 calls will be made to put, get, and remove.
    • 

  • Time Complexity: O(1) average for each operation
    • 

  • Space Complexity: O(n) where n is the number of key-value pairs
    •
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Hint
Solution

Solution Hints

    
  
  • Use an array of linked lists to handle collisions (chaining method).
    • 
  
  • Consider a simple hash function like key % array_size.
    • 
  
  • When handling collisions, search through the linked list to find the matching key.
    • 
  
  • For better performance, choose a prime number for the array size.
    • 
  
  • Consider resizing the array when the load factor exceeds a threshold.
    •

Steps to solve by this approach:

 Step 1: Create a hash function to map keys to bucket indices
 Step 2: Use separate chaining with linked lists to handle collisions
 Step 3: Implement put() to insert or update key-value pairs
 Step 4: Implement get() to retrieve values by key
 Step 5: Implement remove() to delete key-value pairs

Submitted Code :