Collision Handling Using Chaining in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node in the linked list for chaining.

DSA Java

class Node {
    int key;
    int value;
    Node next;

    Node(int key, int value) {
        this.key = key;
        this.value = value;
        this.next = [1];
    }
}

Drag options to blanks, or click blank then click option'

Ahead

Bthis

Cnew Node()

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next to 'this' causes a self-loop.

Using 'new Node()' without parameters causes a compile error.

Assigning next to 'head' before linking is incorrect.

When creating a new node, its next pointer should be null initially because it does not point to any other node yet.

2fill in blank

medium

Complete the code to compute the hash index for a given key.

DSA Java

int getIndex(int key) {
    return key [1] size;
}

Drag options to blanks, or click blank then click option'

A%

B/

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '/' returns a quotient, not an index.

Using '*' or '-' does not limit the index within array bounds.

The modulo operator '%' is used to ensure the index stays within the array size.

3fill in blank

hard

Fix the error in the insert method to correctly add a new node at the beginning of the chain.

DSA Java

void insert(int key, int value) {
    int index = getIndex(key);
    Node newNode = new Node(key, value);
    newNode.next = [1];
    table[index] = newNode;
}

Drag options to blanks, or click blank then click option'

AnewNode

Bnull

Ctable[index]

Dtable

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next to null breaks the chain.

Setting next to newNode causes a self-loop.

Using 'table' without index is invalid.

The new node's next should point to the current head of the chain at that index before updating the head.

4fill in blank

hard

Fill both blanks to complete the search method that finds a value by key in the chain.

DSA Java

int search(int key) {
    int index = getIndex(key);
    Node current = table[[1]];
    while (current != null) {
        if (current.key [2] key) {
            return current.value;
        }
        current = current.next;
    }
    return -1;
}

Drag options to blanks, or click blank then click option'

Aindex

B==

C!=

Dkey

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causes wrong search logic.

Accessing table with 'key' instead of 'index' causes errors.

We start at the chain head at 'index' and check if current.key equals the search key.

5fill in blank

hard

Fill all three blanks to complete the delete method that removes a node by key from the chain.

DSA Java

void delete(int key) {
    int index = getIndex(key);
    Node current = table[index];
    Node prev = null;
    while (current != null) {
        if (current.key [1] key) {
            if (prev == null) {
                table[index] = current.[2];
            } else {
                prev.[3] = current.next;
            }
            return;
        }
        prev = current;
        current = current.next;
    }
}

Drag options to blanks, or click blank then click option'

A==

Bnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' for link updates.

Using '!=' instead of '==' in the if condition.

Not updating table[index] when deleting the first node.

We check if current.key equals key, then update links: if deleting head, update table[index] to current.next; else link prev.next to current.next.

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