Collision Handling Using Chaining in DSA Java

Mental Model

When two keys want to go to the same spot in a table, we keep a small list there to hold both.

Analogy: Imagine a mailbox with several slots. If two letters come for the same slot, you put them both in a small basket inside that slot instead of throwing one away.

Hash Table:
[0] -> 15 -> null
[1] -> null
[2] -> 7 -> 12 -> null
[3] -> null
[4] -> null

Dry Run Walkthrough

Input: Insert keys 7, 12, 15 into a hash table of size 5 using chaining

Goal: Show how collisions are handled by linking keys in the same bucket

Step 1: Insert 7: compute index 7 % 5 = 2, place 7 at bucket 2

[0] -> null
[1] -> null
[2] -> 7 -> null
[3] -> null
[4] -> null

Why: 7 goes to bucket 2 because 7 mod 5 is 2

Step 2: Insert 12: compute index 12 % 5 = 2, add 12 to bucket 2 chain after 7

[0] -> null
[1] -> null
[2] -> 7 -> 12 -> null
[3] -> null
[4] -> null

Why: 12 collides with 7 at bucket 2, so we chain it after 7

Step 3: Insert 15: compute index 15 % 5 = 0, place 15 at bucket 0

[0] -> 15 -> null
[1] -> null
[2] -> 7 -> 12 -> null
[3] -> null
[4] -> null

Why: 15 goes to bucket 0 because 15 mod 5 is 0

Result:

[0] -> 15 -> null
[1] -> null
[2] -> 7 -> 12 -> null
[3] -> null
[4] -> null

Annotated Code

DSA Java

import java.util.LinkedList;

class HashTable {
    private final int SIZE = 5;
    private LinkedList<Integer>[] table;

    @SuppressWarnings("unchecked")
    public HashTable() {
        table = new LinkedList[SIZE];
        for (int i = 0; i < SIZE; i++) {
            table[i] = new LinkedList<>();
        }
    }

    public void insert(int key) {
        int index = key % SIZE; // compute bucket index
        table[index].add(key); // add key to chain at bucket
    }

    public void printTable() {
        for (int i = 0; i < SIZE; i++) {
            System.out.print("[" + i + "] -> ");
            for (int key : table[i]) {
                System.out.print(key + " -> ");
            }
            System.out.println("null");
        }
    }
}

public class Main {
    public static void main(String[] args) {
        HashTable ht = new HashTable();
        ht.insert(7);
        ht.insert(12);
        ht.insert(15);
        ht.printTable();
    }
}

int index = key % SIZE; // compute bucket index

calculate bucket for key using modulo

table[index].add(key); // add key to chain at bucket

append key to linked list at computed bucket

for (int i = 0; i < SIZE; i++) { ... }

iterate over all buckets to print their chains

Output

[0] -> 15 -> null

[1] -> null

[2] -> 7 -> 12 -> null

[3] -> null

[4] -> null

Complexity Analysis

Time: O(1) average for insert because chaining allows quick addition at bucket; worst case O(n) if all keys collide

Space: O(n) because each inserted key is stored in a linked list node

vs Alternative: Compared to open addressing which stores keys directly in array slots, chaining uses extra space for linked lists but handles collisions more flexibly

Edge Cases

Inserting keys that all hash to the same bucket

All keys are stored in one linked list chain at that bucket

DSA Java

table[index].add(key); // add key to chain at bucket

Inserting into an empty hash table

Keys are added normally to empty linked lists

DSA Java

for (int i = 0; i < SIZE; i++) { table[i] = new LinkedList<>(); }

When to Use This Pattern

When you see a hash table with collisions and need to store multiple keys per bucket, use chaining to keep a linked list at each bucket.

Common Mistakes

Mistake: Overwriting existing key in bucket instead of adding to chain
Fix: Always add new keys to the linked list at the bucket instead of replacing

Summary

It stores multiple keys in the same bucket using linked lists to handle collisions.

Use it when hash collisions happen and you want to keep all keys safely.

The key insight is to keep a small list at each bucket to hold all collided keys.

Practice

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

easy

A. Replace the existing element with the new one

B. Resize the hash table immediately on collision

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

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 collision

   Chaining stores all collided elements in a linked list at that index, allowing multiple elements.

3. Final Answer:

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

4. Quick Check:

   Chaining = linked list at index [OK]

Hint: Chaining means linked list at one index for collisions [OK]

Common Mistakes:

• Thinking collision replaces old element
• Assuming resizing happens immediately on collision
• Ignoring collided elements

2. Which of the following is the correct way to declare a hash table with chaining in Java using LinkedList?

easy

A. LinkedList<Integer> table = new LinkedList<>();

B. LinkedList<Integer>[] table = new LinkedList[10];

C. LinkedList<Integer>[] table = new LinkedList<>[10];

D. LinkedList<Integer> table = new LinkedList[10];

Solution

1. Step 1: Understand array of linked lists declaration

   We need an array of LinkedList objects, so use LinkedList<Integer>[].

2. Step 2: Correct Java syntax for array of generic linked lists

   Java allows new LinkedList[10] but not new LinkedList<>[10], so LinkedList<Integer>[] table = new LinkedList[10]; is correct.

3. Final Answer:

   LinkedList<Integer>[] table = new LinkedList[10]; -> Option B

4. Quick Check:

   Array of LinkedList = LinkedList[] table [OK]

Hint: Array of LinkedList uses LinkedList<Type>[] [OK]

Common Mistakes:

• Trying to use new LinkedList<>[size]
• Declaring LinkedList without array
• Wrong generic syntax in array creation

3. Consider the following Java code snippet for a hash table with chaining:

import java.util.LinkedList;

public class HashTable {
    LinkedList<Integer>[] table;
    int size = 5;

    public HashTable() {
        table = new LinkedList[size];
        for (int i = 0; i < size; i++) {
            table[i] = new LinkedList<>();
        }
    }

    public void insert(int key) {
        int index = key % size;
        table[index].add(key);
    }

    public void printTable() {
        for (int i = 0; i < size; i++) {
            System.out.print(i + ": ");
            for (int val : table[i]) {
                System.out.print(val + " -> ");
            }
            System.out.println("null");
        }
    }

    public static void main(String[] args) {
        HashTable ht = new HashTable();
        ht.insert(7);
        ht.insert(12);
        ht.insert(17);
        ht.printTable();
    }
}

What is the output of this program?

medium

A. 0: null
1: null
2: null
3: 7 -> 12 -> 17 -> null
4: null

B. 0: 7 -> 12 -> 17 -> null
1: null
2: null
3: null
4: null

C. 0: 12 -> null
1: null
2: 7 -> 17 -> null
3: null
4: null

D. 0: null
1: null
2: 7 -> 12 -> 17 -> null
3: null
4: null

Solution

1. Step 1: Calculate index for each inserted key

   Size is 5, so index = key % 5. 7 % 5 = 2, 12 % 5 = 2, 17 % 5 = 2.

2. Step 2: Insert keys at index 2 and print table

   All keys go to table[2]. So index 2 has 7 -> 12 -> 17 -> null, others are empty.

3. Final Answer:

   0: null 1: null 2: 7 -> 12 -> 17 -> null 3: null 4: null -> Option D

4. Quick Check:

   Keys mod 5 = 2, all at index 2 [OK]

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

Common Mistakes:

• Miscomputing modulo operation
• Assuming keys go to different indices
• Confusing index with key value

4. The following code snippet tries to insert keys into a hash table with chaining but throws a NullPointerException. What is the error?

import java.util.LinkedList;

public class HashTable {
    LinkedList<Integer>[] table = new LinkedList[5];

    public void insert(int key) {
        int index = key % 5;
        table[index].add(key);
    }

    public static void main(String[] args) {
        HashTable ht = new HashTable();
        ht.insert(10);
    }
}

medium

A. The array of LinkedLists is not initialized with LinkedList objects

B. The modulo operation is incorrect

C. The insert method should return a value

D. The LinkedList should be replaced with ArrayList

Solution

1. Step 1: Identify cause of NullPointerException

   The array 'table' is created but each element is null because LinkedList objects are not initialized.

2. Step 2: Fix by initializing each LinkedList in the array

   Before inserting, each table[i] must be assigned new LinkedList<>() to avoid null pointer.

3. Final Answer:

   The array of LinkedLists is not initialized with LinkedList objects -> Option A

4. Quick Check:

   NullPointerException = uninitialized LinkedList array [OK]

Hint: Initialize each LinkedList before use [OK]

Common Mistakes:

• Ignoring initialization of linked lists
• Blaming modulo operation
• Thinking insert needs return value

5. You want to implement a hash table with chaining that supports insert and search operations. Given the following partial code, fill in the missing search method to return true if the key exists, otherwise false.

import java.util.LinkedList;

public class HashTable {
    LinkedList<Integer>[] table;
    int size = 7;

    public HashTable() {
        table = new LinkedList[size];
        for (int i = 0; i < size; i++) {
            table[i] = new LinkedList<>();
        }
    }

    public void insert(int key) {
        int index = key % size;
        table[index].add(key);
    }

    // Fill this method
    public boolean search(int key) {
        // ???
    }
}

Which of the following is the correct implementation of the search method?

hard

A. int index = key % size;
return table[index].contains(key);

B. int index = key % size;
return table[index].get(0) == key;

C. for (LinkedList<Integer> list : table) {
if (list.contains(key)) return true;
}
return false;

D. int index = key % size;
return table[index].indexOf(key) == -1;

Solution

1. Step 1: Calculate index using modulo

   Use key % size to find the correct linked list index.

2. Step 2: Check if linked list at index contains the key

   Use contains() method on linked list to check presence of key.

3. Final Answer:

   int index = key % size; return table[index].contains(key); -> Option A

4. Quick Check:

   Search = contains at index [OK]

Hint: Search key in linked list at key % size [OK]

Common Mistakes:

• Checking only first element in list
• Searching all lists instead of one
• Using indexOf == -1 incorrectly