HashMap Implementation from Scratch in DSA Python - Time & Space Complexity

Time Complexity: HashMap Implementation from Scratch

O(n) worst case, O(1) average case

Understanding Time Complexity

When building a HashMap from scratch, it's important to understand how fast it can add, find, or remove items.

We want to know how the time to do these actions changes as we store more items.

Scenario Under Consideration

Analyze the time complexity of the following simple HashMap put operation.

class SimpleHashMap:
    def __init__(self, size=10):
        self.size = size
        self.buckets = [[] for _ in range(size)]

    def put(self, key, value):
        index = hash(key) % self.size
        bucket = self.buckets[index]
        for i, (k, v) in enumerate(bucket):
            if k == key:
                bucket[i] = (key, value)
                return
        bucket.append((key, value))

This code stores key-value pairs in buckets using a hash function and handles collisions by storing multiple pairs in a list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through the bucket list to find if the key exists.
• How many times: Up to the number of items in that bucket, which depends on how many keys hash to the same index.

How Execution Grows With Input

As more items are added, buckets may hold more pairs, making the search inside a bucket longer.

Input Size (n)	Approx. Operations
10	About 1 to 2 checks per put (few collisions)
100	About 5 to 10 checks per put (more collisions)
1000	About 50 to 100 checks per put (many collisions)

Pattern observation: Without resizing, the time grows roughly proportional to the number of items per bucket, which grows as total items increase.

Final Time Complexity

Time Complexity: O(n) in worst case, O(1) on average

This means usually adding or updating a key is very fast, but if many keys collide, it can slow down linearly with the number of items.

Common Mistake

[X] Wrong: "HashMap operations always take the same constant time no matter what."

[OK] Correct: If many keys end up in the same bucket, the code must check each one, making the operation slower.

Interview Connect

Understanding how collisions affect speed shows you know the real behavior of hash maps, a key skill for coding interviews.

Self-Check

"What if we resize and rehash the buckets when they get too full? How would the time complexity change then?"

Practice

1. What is the main purpose of a HashMap in data structures?

easy

A. To store key-value pairs for fast data retrieval

B. To store data in a sorted order

C. To perform mathematical calculations

D. To store data in a sequential list

Solution

1. Step 1: Understand what a HashMap stores

   A HashMap stores data as key-value pairs, allowing quick access to values using keys.

2. Step 2: Identify the main purpose

   The main purpose is to retrieve data quickly by keys, not to sort or calculate.

3. Final Answer:

   To store key-value pairs for fast data retrieval -> Option A

4. Quick Check:

   HashMap = fast key-value access [OK]

Hint: Remember: HashMap = key-value pairs for quick lookup [OK]

Common Mistakes:

• Confusing HashMap with sorted data structures
• Thinking HashMap stores data sequentially
• Assuming HashMap is for calculations

2. Which of the following is the correct way to define a simple HashMap class constructor in Python?

easy

A. def __init__(self): self.map = set()

B. def __init__(self): self.map = []

C. def __init__(self): self.map = {}

D. def __init__(self): self.map = ()

Solution

1. Step 1: Identify the data structure for storing key-value pairs

   A HashMap stores key-value pairs, so a dictionary {} is appropriate.

2. Step 2: Check each option's data type

   def __init__(self): self.map = {} uses {} which is a dictionary. Others use list, set, or tuple which are not suitable for key-value mapping.

3. Final Answer:

   def __init__(self): self.map = {} -> Option C

4. Quick Check:

   HashMap uses dictionary = {} [OK]

Hint: Use {} to store key-value pairs in Python [OK]

Common Mistakes:

• Using list or tuple instead of dictionary
• Confusing set with dictionary
• Not initializing the map attribute

3. What will be the output of the following code?

class HashMap:
    def __init__(self):
        self.map = {}
    def put(self, key, value):
        self.map[key] = value
    def get(self, key):
        return self.map.get(key, None)

h = HashMap()
h.put('a', 1)
h.put('b', 2)
print(h.get('a'))
print(h.get('c'))

medium

A. 1\nNone

B. 1\n0

C. None\nNone

D. Error

Solution

1. Step 1: Trace the put method calls

   Keys 'a' and 'b' are added with values 1 and 2 respectively.

2. Step 2: Trace the get method calls

   Getting 'a' returns 1. Getting 'c' returns None because 'c' is not in the map and default is None.

3. Final Answer:

   1\nNone -> Option A

4. Quick Check:

   Existing key returns value, missing key returns None [OK]

Hint: get() returns None if key not found by default [OK]

Common Mistakes:

• Expecting 0 instead of None for missing keys
• Confusing print output format
• Assuming error on missing key

4. Find the error in this HashMap implementation snippet:

class HashMap:
    def __init__(self):
        self.map = []
    def put(self, key, value):
        self.map[key] = value

medium

A. Missing return statement in put method

B. No error, code is correct

C. Incorrect method name put instead of add

D. Using list instead of dictionary for self.map

Solution

1. Step 1: Check the data structure used for self.map

   self.map is initialized as a list, but lists do not support key-based assignment.

2. Step 2: Understand why list causes error

   Using self.map[key] = value will cause TypeError because list indices must be integers, not keys.

3. Final Answer:

   Using list instead of dictionary for self.map -> Option D

4. Quick Check:

   List can't use keys as indices [OK]

Hint: Use dictionary {} for key-value storage, not list [] [OK]

Common Mistakes:

• Thinking put method needs return
• Confusing method names
• Ignoring data type of self.map

5. You want to implement a HashMap that handles collisions by chaining (using lists). Which of the following code snippets correctly initializes the internal storage for chaining?

hard

A. self.map = {}

B. self.map = [[] for _ in range(size)]

C. self.map = [None] * size

D. self.map = [0] * size

Solution

1. Step 1: Understand chaining collision handling

   Chaining uses a list of lists where each bucket holds a list of key-value pairs.

2. Step 2: Identify correct initialization

   self.map = [[] for _ in range(size)] creates a list of empty lists, one for each bucket, suitable for chaining.

3. Step 3: Why other options are incorrect

   self.map = [None] * size creates a list of None, which needs extra checks. self.map = {} is a dictionary, not a list of buckets. self.map = [0] * size creates a list of zeros, not suitable for chaining.

4. Final Answer:

   self.map = [[] for _ in range(size)] -> Option B

5. Quick Check:

   Chaining needs list of lists = [[] for _ in range(size)] [OK]

Hint: Use list comprehension to create buckets for chaining [OK]

Common Mistakes:

• Using None or 0 instead of empty lists for buckets
• Confusing dictionary with list of buckets
• Not initializing buckets properly for chaining