Collision Handling Using Chaining in DSA Python - Time & Space Complexity

Time Complexity: Collision Handling Using Chaining

O(1 + n/k)

Understanding Time Complexity

When we use chaining to handle collisions in a hash table, we want to know how fast operations like search, insert, and delete run.

We ask: How does the time to do these operations grow as the number of items increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class HashTable:
    def __init__(self, size):
        self.size = size
        self.table = [[] for _ in range(size)]

    def insert(self, key, value):
        index = hash(key) % self.size
        self.table[index].append((key, value))

    def search(self, key):
        index = hash(key) % self.size
        for k, v in self.table[index]:
            if k == key:
                return v
        return None

This code shows a hash table using chaining where each bucket holds a list of key-value pairs.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through the linked list (chain) at a bucket during search or insert.
• How many times: Depends on the number of items in that bucket, which grows with collisions.

How Execution Grows With Input

As more items are added, chains get longer if the table size stays the same.

Input Size (n)	Approx. Operations (search in chain)
10	About 1 (if table size ~10, chains short)
100	About 10 (chains longer if table size unchanged)
1000	About 100 (chains much longer without resizing)

Pattern observation: If the table size stays fixed, operations grow roughly proportional to the number of items per bucket.

Final Time Complexity

Time Complexity: O(1 + n/k)

This means the time depends on the average chain length, which is total items divided by table size.

Common Mistake

[X] Wrong: "Hash table operations are always O(1) no matter what."

[OK] Correct: If many items collide into the same bucket, the chain grows and operations slow down, making time depend on chain length.

Interview Connect

Understanding how chaining affects time helps you explain hash table performance clearly and shows you know how to handle collisions well.

Self-Check

"What if we resize the hash table when chains get too long? How would that change the time complexity?"

Practice

1. What is the main idea behind collision handling using chaining in hash tables?

easy

A. Store collided elements in a linked list at the same index

B. Replace the existing element with the new one

C. Resize the hash table immediately

D. Ignore the new element causing collision

Solution

1. Step 1: Understand collision in hash tables

   When two keys hash to the same index, a collision occurs.

2. Step 2: How chaining handles collisions

   Chaining stores all collided elements in a linked list at that index, allowing multiple elements to share the same slot.

3. Final Answer:

   Store collided elements in a linked list at the same index -> Option A

4. Quick Check:

   Chaining = linked list at index [OK]

Hint: Remember: chaining uses linked lists to store collisions [OK]

Common Mistakes:

• Thinking collisions replace existing elements
• Assuming collisions are ignored
• Believing resizing happens immediately on collision

2. Which of the following is the correct way to initialize a hash table with chaining in Python?

easy

A. hash_table = dict()

B. hash_table = [None] * size

C. hash_table = [0] * size

D. hash_table = [[] for _ in range(size)]

Solution

1. Step 1: Understand chaining initialization

   Chaining requires each slot to hold a list to store multiple elements.

2. Step 2: Choose correct initialization

   Using list comprehension to create a list of empty lists ensures each slot can hold multiple elements.

3. Final Answer:

   hash_table = [[] for _ in range(size)] -> Option D

4. Quick Check:

   Empty lists per slot = hash_table = [[] for _ in range(size)] [OK]

Hint: Use list of empty lists for chaining slots [OK]

Common Mistakes:

• Using [None] * size creates references to same list
• Using zeros instead of lists for chaining
• Using dict() which is not a list-based hash table

3. Consider the following code using chaining for collision handling:

hash_table = [[] for _ in range(3)]
keys = [5, 8, 11]
for key in keys:
    index = key % 3
    hash_table[index].append(key)
print(hash_table)

What is the output?

medium

A. [[5, 8, 11], [], []]

B. [[], [], [5, 8, 11]]

C. [[], [5, 8, 11], []]

D. [[], [5], [8, 11]]

Solution

1. Step 1: Calculate indices for each key

   5 % 3 = 2, 8 % 3 = 2, 11 % 3 = 2. All keys map to index 2.

2. Step 2: Append keys to corresponding lists

   All keys 5, 8, 11 are appended to hash_table[2], so hash_table = [[], [], [5, 8, 11]].

3. Final Answer:

   [[], [], [5, 8, 11]] -> Option B

4. Quick Check:

   Keys mod 3 = 2, all in last list [OK]

Hint: Calculate key % size to find index, then append [OK]

Common Mistakes:

• Miscomputing modulo operation
• Appending keys to wrong index
• Confusing list positions in output

4. Find the error in this chaining insertion code:

hash_table = [[]] * 4
key = 10
index = key % 4
hash_table[index].append(key)
print(hash_table)

medium

A. Modulo operation is incorrect

B. Append method is not valid for lists

C. Using [[]] * 4 creates references to the same list

D. Index calculation should be key % 3

Solution

1. Step 1: Understand list multiplication behavior

   Using [[]] * 4 creates 4 references to the same inner list, not 4 separate lists.

2. Step 2: Effect on append operation

   Appending to one index appends to all because all slots point to the same list.

3. Final Answer:

   Using [[]] * 4 creates references to the same list -> Option C

4. Quick Check:

   List multiplication duplicates references [OK]

Hint: Use list comprehension for separate lists, not multiplication [OK]

Common Mistakes:

• Assuming [[]] * 4 creates independent lists
• Incorrect modulo calculation
• Misunderstanding append method usage

5. You have a hash table of size 5 using chaining. Insert keys [7, 12, 17, 22, 27]. After insertion, what is the structure of the hash table?

hard

A. [[], [], [7, 12, 17, 22, 27], [], []]

B. [[], [7, 12, 17, 22, 27], [], [], []]

C. [[], [], [], [7, 12, 17, 22, 27], []]

D. [[7, 12, 17, 22, 27], [], [], [], []]

Solution

1. Step 1: Calculate index for each key using modulo 5

   7 % 5 = 2, 12 % 5 = 2, 17 % 5 = 2, 22 % 5 = 2, 27 % 5 = 2. All keys map to index 2.

2. Step 2: Insert keys at index 2

   All keys are appended to the list at index 2, so hash_table = [[], [], [7, 12, 17, 22, 27], [], []].

3. Final Answer:

   [[], [], [7, 12, 17, 22, 27], [], []] -> Option A

4. Quick Check:

   All keys mod 5 = 2, stored at index 2 [OK]

Hint: Calculate key % size to find correct bucket [OK]

Common Mistakes:

• Miscomputing modulo results
• Placing keys in wrong index
• Confusing list positions in output