Collision Handling Using Open Addressing Linear Probing in DSA C# - Build from Scratch

Collision Handling Using Open Addressing Linear Probing

📖 Scenario: Imagine you are building a simple phone book app that stores phone numbers by name. You want to store names and phone numbers in a small table. Sometimes, two names might want to use the same spot in the table. To handle this, you will use a method called linear probing to find the next free spot.

🎯 Goal: You will create a small hash table using an array. You will add names and phone numbers. When a spot is taken, you will use linear probing to find the next empty spot. Finally, you will print the table showing where each name and number is stored.

📋 What You'll Learn

Create an array to hold phone book entries with fixed size 5

Create a simple hash function to find index from name

Use linear probing to handle collisions when inserting entries

Print the final state of the phone book array showing index, name, and phone number

💡 Why This Matters

🌍 Real World

Hash tables with collision handling are used in databases, caches, and many software systems to store and find data quickly.

💼 Career

Understanding collision handling and hash tables is important for software developers working on performance-critical applications and data storage.

1

Create the phone book array

Create a class called Entry with two public string fields: Name and PhoneNumber. Then create an array called phoneBook of type Entry with size 5.

DSA C#

// Define Entry class and create phoneBook array
// Your code here

Hint

Define a class with two string fields. Then create an array of that class with size 5.

2

Create the hash function

Create a function called HashFunction that takes a string name and returns an integer index. The function should return the sum of the ASCII values of the characters in name modulo 5.

DSA C#

public class Entry
{
    public string Name;
    public string PhoneNumber;
}

Entry[] phoneBook = new Entry[5];

// Create HashFunction method
// Your code here

Hint

Loop over each character in the name, add its ASCII value to a sum, then return sum modulo 5.

3

Insert entries using linear probing

Create a method called InsertEntry that takes a string name and string phoneNumber. Use HashFunction to find the starting index. If the spot is empty, insert a new Entry. If not, use linear probing by checking the next index (wrap around using modulo 5) until an empty spot is found.

DSA C#

public class Entry
{
    public string Name;
    public string PhoneNumber;
}

Entry[] phoneBook = new Entry[5];

int HashFunction(string name)
{
    int sum = 0;
    foreach (char c in name)
    {
        sum += c;
    }
    return sum % 5;
}

// Create InsertEntry method
// Your code here

Hint

Use a loop to check if the spot is taken. If yes, move to next index using modulo. Insert when empty spot found.

4

Print the phone book

Use a for loop with variable i from 0 to 4 to print each index of phoneBook. If the entry is not null, print Index i: Name - PhoneNumber. If null, print Index i: Empty.

DSA C#

public class Entry
{
    public string Name;
    public string PhoneNumber;
}

Entry[] phoneBook = new Entry[5];

int HashFunction(string name)
{
    int sum = 0;
    foreach (char c in name)
    {
        sum += c;
    }
    return sum % 5;
}

void InsertEntry(string name, string phoneNumber)
{
    int index = HashFunction(name);
    int startIndex = index;
    while (phoneBook[index] != null)
    {
        index = (index + 1) % 5;
        if (index == startIndex)
        {
            // Table full, cannot insert
            return;
        }
    }
    phoneBook[index] = new Entry { Name = name, PhoneNumber = phoneNumber };
}

// Insert sample entries
InsertEntry("Alice", "12345");
InsertEntry("Bob", "23456");
InsertEntry("Charlie", "34567");
InsertEntry("David", "45678");
InsertEntry("Eve", "56789");

// Print the phone book
for (int i = 0; i < 5; i++)
{
    // Your code here
}

Hint

Use a for loop from 0 to 4. Check if phoneBook[i] is null. Print name and number if not null, else print 'Empty'.

Practice

1. What is the main idea behind Linear Probing in collision handling for hash tables?

easy

A. Use a linked list at each slot to store multiple elements.

B. If a collision occurs, check the next slot sequentially until an empty slot is found.

C. Double the size of the hash table when a collision happens.

D. Replace the existing element in the slot with the new one.

Solution

1. Step 1: Understand collision handling methods

   Linear probing is a method where if a collision happens, the algorithm checks the next slot in the array sequentially.

2. Step 2: Compare options with linear probing definition

   If a collision occurs, check the next slot sequentially until an empty slot is found. correctly describes checking the next slot sequentially until an empty slot is found, which matches linear probing.

3. Final Answer:

   If a collision occurs, check the next slot sequentially until an empty slot is found. -> Option B

4. Quick Check:

   Linear probing = check next slot sequentially [OK]

Hint: Linear probing means try next slot one by one [OK]

Common Mistakes:

• Confusing linear probing with chaining
• Thinking linear probing replaces existing elements
• Assuming table size doubles on collision

2. Which of the following is the correct way to calculate the next index in linear probing for a hash table of size tableSize when a collision occurs at index?

easy

A. nextIndex = (index * 2) % tableSize;

B. nextIndex = (index - 1) % tableSize;

C. nextIndex = (index + 1) % tableSize;

D. nextIndex = (index + 3) % tableSize;

Solution

1. Step 1: Recall linear probing formula

   Linear probing moves to the next slot by adding 1 to the current index and taking modulo with table size to wrap around.

2. Step 2: Match formula with options

   nextIndex = (index + 1) % tableSize; uses (index + 1) % tableSize, which is the correct formula for linear probing.

3. Final Answer:

   nextIndex = (index + 1) % tableSize; -> Option C

4. Quick Check:

   Linear probing next index = (index + 1) % size [OK]

Hint: Next index = (current index + 1) mod table size [OK]

Common Mistakes:

• Multiplying index instead of adding
• Subtracting instead of adding
• Adding a number other than 1

3. Consider a hash table of size 5 using linear probing. The hash function is h(key) = key % 5. Insert keys in this order: 7, 12, 17. What will be the final state of the hash table (index: value)?

medium

A. {0: null, 1: null, 2: 7, 3: 12, 4: 17}

B. {0: 12, 1: 17, 2: 7, 3: null, 4: null}

C. {0: null, 1: 7, 2: null, 3: 12, 4: 17}

D. {0: null, 1: 7, 2: 17, 3: 12, 4: null}

Solution

1. Step 1: Calculate initial indices for each key

   7 % 5 = 2, 12 % 5 = 2, 17 % 5 = 2. All keys initially map to index 2.

2. Step 2: Insert keys using linear probing

   Insert 7 at index 2 (empty). Insert 12: index 2 occupied, try index 3 (empty), insert 12. Insert 17: index 2 and 3 occupied, try index 4 (empty), insert 17.

3. Final Answer:

   {0: null, 1: null, 2: 7, 3: 12, 4: 17} -> Option A

4. Quick Check:

   Keys collide at 2, placed at 2,3,4 sequentially [OK]

Hint: All keys hash to same index, place next free slot [OK]

Common Mistakes:

• Placing all keys at index 2 ignoring collisions
• Not wrapping around the table
• Inserting keys in wrong order

4. The following C# code tries to insert keys into a hash table using linear probing. What is the bug?

int tableSize = 5;
int[] hashTable = new int[tableSize];
for (int i = 0; i < tableSize; i++) hashTable[i] = -1;

int Hash(int key) => key % tableSize;

void Insert(int key) {
    int index = Hash(key);
    while (hashTable[index] != -1) {
        index = (index + 1);
    }
    hashTable[index] = key;
}

Insert(7);
Insert(12);
Insert(17);

medium

A. The hash function is incorrect for linear probing.

B. The Insert method should use recursion instead of a loop.

C. The initial hash table values should be 0, not -1.

D. The code does not wrap index back to 0 when it reaches the end of the table.

Solution

1. Step 1: Analyze the index update in the loop

   The code increments index by 1 on collision but does not use modulo to wrap around the table size.

2. Step 2: Identify the consequence of missing modulo

   Without modulo, index can go beyond array bounds causing runtime error or incorrect behavior.

3. Final Answer:

   The code does not wrap index back to 0 when it reaches the end of the table. -> Option D

4. Quick Check:

   Linear probing requires modulo to wrap index [OK]

Hint: Always use modulo to wrap index in linear probing [OK]

Common Mistakes:

• Forgetting modulo causes index out of range
• Changing hash function unnecessarily
• Thinking initial values must be zero

5. You have a hash table of size 7 using linear probing. The hash function is h(key) = key % 7. You insert keys: 10, 24, 31, 17, 3. Then you delete key 24 by marking its slot as empty. Now you want to search for key 31. What problem might occur and how to fix it?

hard

A. Search may stop early at the empty slot of deleted key 24, missing key 31; fix by using a special 'deleted' marker instead of empty.

B. Search will always find key 31 without issues; no fix needed.

C. Key 31 will be overwritten by key 24; fix by rehashing all keys after deletion.

D. The hash function is wrong; fix by changing it to key % 5.

Solution

1. Step 1: Understand deletion effect in linear probing

   Deleting by marking slot empty breaks the probe chain, so search may stop early and miss keys inserted after the deleted slot.

2. Step 2: Identify correct deletion method

   Use a special 'deleted' marker to indicate slot is free for insertion but search should continue probing.

3. Final Answer:

   Search may stop early at the empty slot of deleted key 24, missing key 31; fix by using a special 'deleted' marker instead of empty. -> Option A

4. Quick Check:

   Deleted slots need special marker to continue search [OK]

Hint: Use 'deleted' marker, not empty slot, to avoid search errors [OK]

Common Mistakes:

• Marking deleted slots as empty breaks search
• Assuming search always finds keys after deletion
• Changing hash function unnecessarily