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

Time Complexity: Collision Handling Using Chaining

O(n)

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.

import java.util.LinkedList;

class HashTable {
  LinkedList[] table;

  int hash(int key) {
    return key % table.length;
  }

  void insert(int key) {
    int index = hash(key);
    table[index].add(key); // add to linked list at index
  }

  boolean search(int key) {
    int index = hash(key);
    return table[index].contains(key); // search linked list
  }
}
    

This code inserts and searches keys in a hash table using chaining, where each bucket holds a linked list of keys.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing the linked list at a hash bucket during search or insert.
• How many times: Depends on the number of keys in that bucket, which varies with collisions.

How Execution Grows With Input

As more keys are added, some buckets get longer linked lists, so searching or inserting takes longer.

Input Size (n)	Approx. Operations
10	About 1 to 2 steps per search (few collisions)
100	About 5 to 10 steps per search (more collisions)
1000	About 50 to 100 steps per search (many collisions)

Pattern observation: If the number of buckets stays fixed, the linked lists grow longer roughly proportional to n, so operations slow down linearly.

Final Time Complexity

Time Complexity: O(n/k) where k is number of buckets, often simplified to O(n) in worst case

This means if many keys collide into the same bucket, operations take longer, growing linearly with the number of keys in that bucket.

Common Mistake

[X] Wrong: "Hash table operations always take constant time, no matter what."

[OK] Correct: If many keys collide into the same bucket, the linked list grows, making search or insert slower than constant time.

Interview Connect

Understanding how chaining affects time complexity helps you explain hash table performance clearly and shows you know how collisions impact speed in real systems.

Self-Check

"What if we used balanced trees instead of linked lists for chaining? How would the time complexity change?"

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