Collision Handling Using Chaining in DSA Python

Mental Model

When two keys want to go to the same spot in a table, we keep them in a small linked list at that spot.

Analogy: Imagine a mailbox with several slots. If two letters come for the same slot, we put them in a small pile inside that slot instead of losing one.

Hash Table:
Index 0 -> null
Index 1 -> 21 -> 41 -> null
Index 2 -> 32 -> null
Index 3 -> null
Index 4 -> 14 -> null

Dry Run Walkthrough

Input: Insert keys 14, 21, 32, 41 into a hash table of size 5 using chaining

Goal: Show how collisions are handled by adding nodes to linked lists at the same index

Step 1: Insert 14, hash index = 14 % 5 = 4

Index 4 -> 14 -> null

Why: 14 goes to index 4, no collision yet

Step 2: Insert 21, hash index = 21 % 5 = 1

Index 1 -> 21 -> null
Index 4 -> 14 -> null

Why: 21 goes to index 1, no collision yet

Step 3: Insert 32, hash index = 32 % 5 = 2

Index 1 -> 21 -> null
Index 2 -> 32 -> null
Index 4 -> 14 -> null

Why: 32 goes to index 2, no collision yet

Step 4: Insert 41, hash index = 41 % 5 = 1, collision with 21

Index 1 -> 21 -> 41 -> null
Index 2 -> 32 -> null
Index 4 -> 14 -> null

Why: 41 collides at index 1, so we add it to the linked list there

Result:

Index 0 -> null
Index 1 -> 21 -> 41 -> null
Index 2 -> 32 -> null
Index 3 -> null
Index 4 -> 14 -> null

Annotated Code

DSA Python

class Node:
    def __init__(self, key):
        self.key = key
        self.next = None

class HashTable:
    def __init__(self, size):
        self.size = size
        self.table = [None] * size

    def hash_function(self, key):
        return key % self.size

    def insert(self, key):
        index = self.hash_function(key)  # find index
        new_node = Node(key)  # create new node
        if self.table[index] is None:
            self.table[index] = new_node  # no collision, place node
        else:
            curr = self.table[index]
            while curr.next is not None:
                curr = curr.next  # move to end of chain
            curr.next = new_node  # add new node at end

    def __str__(self):
        result = []
        for i in range(self.size):
            curr = self.table[i]
            chain = []
            while curr:
                chain.append(str(curr.key))
                curr = curr.next
            chain_str = ' -> '.join(chain) + ' -> null' if chain else 'null'
            result.append(f'Index {i} -> {chain_str}')
        return '\n'.join(result)

# Driver code
ht = HashTable(5)
keys = [14, 21, 32, 41]
for key in keys:
    ht.insert(key)
print(ht)

index = self.hash_function(key) # find index

compute hash index for the key

if self.table[index] is None:

check if slot is empty, no collision

curr = self.table[index] while curr.next is not None: curr = curr.next

traverse linked list to find last node

curr.next = new_node # add new node at end

append new node to handle collision

Output

Index 0 -> null

Index 1 -> 21 -> 41 -> null

Index 2 -> 32 -> null

Index 3 -> null

Index 4 -> 14 -> null

Complexity Analysis

Time: O(n) in worst case because all keys could hash to the same index forming a long chain

Space: O(n) because we store all keys in linked lists inside the table

vs Alternative: Compared to open addressing which stores all keys in the array itself, chaining uses extra linked lists but handles collisions more flexibly

Edge Cases

Inserting into an empty hash table

Key is placed directly without collision

DSA Python

if self.table[index] is None:

Multiple keys hashing to same index

Keys are chained in linked list at that index

DSA Python

while curr.next is not None:

Hash table size is 1 (all keys collide)

All keys form one long chain at index 0

DSA Python

while curr.next is not None:

When to Use This Pattern

When you see a hash table with collisions and need to store multiple keys at one index, use chaining to keep keys in a linked list at that index.

Common Mistakes

Mistake: Overwriting existing key at index instead of chaining
Fix: Traverse the linked list and add new node at the end instead of replacing

Mistake: Not handling empty slot before chaining
Fix: Check if slot is None and insert directly if so

Summary

It stores multiple keys at the same hash index using linked lists to handle collisions.

Use it when your hash function causes collisions and you want to keep all keys safely.

The key insight is to keep a small chain of keys at each index instead of losing or overwriting them.