What if your keys always got lost in the same drawer? Chaining is the magic that keeps them all in order!
Why Collision Handling Using Chaining in DSA Java?
Imagine you have a big box with many small compartments to store your keys. But sometimes, two keys end up in the same compartment because they look similar. You try to find your key by checking each key in that compartment one by one.
Manually checking each key in the same compartment is slow and confusing. If many keys share the same spot, you waste time searching and might even lose track of which keys you already checked.
Collision Handling Using Chaining organizes keys that share the same compartment into a small linked list. This way, you keep all keys neatly connected and can quickly find or add keys without mixing them up.
int index = hash(key); if (array[index] == null) { array[index] = key; } else { // no structure to handle collision // keys overwrite or lost }
int index = hash(key);
if (array[index] == null) {
array[index] = new LinkedList<>();
}
array[index].add(key);This method lets us store many items efficiently even when they share the same spot, making data retrieval fast and reliable.
Think of a library where books with the same category are placed on the same shelf but arranged in a chain so you can easily find the exact book you want.
Manual collision handling is slow and error-prone.
Chaining uses linked lists to keep collided items organized.
This improves search and insertion speed in hash tables.
Practice
collision handling using chaining in hash tables?Solution
Step 1: Understand collision in hash tables
When two keys hash to the same index, a collision occurs.Step 2: How chaining handles collision
Chaining stores all collided elements in a linked list at that index, allowing multiple elements.Final Answer:
Store collided elements in a linked list at the same index -> Option CQuick Check:
Chaining = linked list at index [OK]
- Thinking collision replaces old element
- Assuming resizing happens immediately on collision
- Ignoring collided elements
LinkedList?Solution
Step 1: Understand array of linked lists declaration
We need an array of LinkedList objects, so use LinkedList<Integer>[].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.Final Answer:
LinkedList<Integer>[] table = new LinkedList[10]; -> Option BQuick Check:
Array of LinkedList = LinkedList[] table [OK]
- Trying to use new LinkedList<>[size]
- Declaring LinkedList without array
- Wrong generic syntax in array creation
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?Solution
Step 1: Calculate index for each inserted key
Size is 5, so index = key % 5. 7 % 5 = 2, 12 % 5 = 2, 17 % 5 = 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.Final Answer:
0: null 1: null 2: 7 -> 12 -> 17 -> null 3: null 4: null -> Option DQuick Check:
Keys mod 5 = 2, all at index 2 [OK]
- Miscomputing modulo operation
- Assuming keys go to different indices
- Confusing index with key value
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);
}
}Solution
Step 1: Identify cause of NullPointerException
The array 'table' is created but each element is null because LinkedList objects are not initialized.Step 2: Fix by initializing each LinkedList in the array
Before inserting, each table[i] must be assigned new LinkedList<>() to avoid null pointer.Final Answer:
The array of LinkedLists is not initialized with LinkedList objects -> Option AQuick Check:
NullPointerException = uninitialized LinkedList array [OK]
- Ignoring initialization of linked lists
- Blaming modulo operation
- Thinking insert needs return value
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?Solution
Step 1: Calculate index using modulo
Use key % size to find the correct linked list index.Step 2: Check if linked list at index contains the key
Use contains() method on linked list to check presence of key.Final Answer:
int index = key % size; return table[index].contains(key); -> Option AQuick Check:
Search = contains at index [OK]
- Checking only first element in list
- Searching all lists instead of one
- Using indexOf == -1 incorrectly
Ready to prove it?
Pass the test and this pattern moves unknown
