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

Collision Handling Using Open Addressing Linear Probing

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

🎯 Goal: Build a hash table using open addressing with linear probing to handle collisions. You will insert entries and then display the final state of the hash table.

📋 What You'll Learn

Create a hash table array of fixed size 7

Insert key-value pairs using a simple hash function

Handle collisions using linear probing

Print the final hash table showing index and stored key-value pairs

💡 Why This Matters

🌍 Real World

Hash tables are used in databases, caches, and many applications to store and quickly find data. Handling collisions efficiently is important for performance.

💼 Career

Understanding collision handling with linear probing is fundamental for software engineers working with data storage, system design, and performance optimization.

1

Create the hash table array

Create a slice called hashTable of size 7 that will hold strings representing stored entries. Initialize all entries with empty strings "".

DSA Go

# Create the hashTable slice with size 7 and initialize with empty strings
# Your code here

Hint

Use make([]string, 7) to create the slice and a for loop to set each element to an empty string.

2

Create the hash function

Create a function called hash that takes a string key and returns an int. The function should return the sum of the ASCII values of the characters in key modulo 7.

DSA Go

package main

import "fmt"

func main() {
    hashTable := make([]string, 7)
    for i := range hashTable {
        hashTable[i] = ""
    }
}

// Create the hash function below
# Your code here

Hint

Loop over each character in the string, add its integer value to a sum, then return sum modulo 7.

3

Insert entries using linear probing

Insert these key-value pairs into hashTable using linear probing to handle collisions: "Bob:123", "Amy:456", "Eve:789". Use the hash function on the key (name before colon) to find the initial index. If the spot is taken, move to the next index until an empty spot is found.

DSA Go

package main

import "fmt"

func hash(key string) int {
    sum := 0
    for _, ch := range key {
        sum += int(ch)
    }
    return sum % 7
}

func main() {
    hashTable := make([]string, 7)
    for i := range hashTable {
        hashTable[i] = ""
    }

    // Insert entries below
    // Your code here
}

Hint

Extract the key before the colon, compute its hash, then check if the spot is empty. If not, move to the next index until you find an empty spot.

4

Print the final hash table

Print the contents of hashTable showing each index and the stored entry. Use a for loop with index i and print in the format: Index i: entry. If the entry is empty, print Index i: empty.

DSA Go

package main

import "fmt"

func hash(key string) int {
    sum := 0
    for _, ch := range key {
        sum += int(ch)
    }
    return sum % 7
}

func main() {
    hashTable := make([]string, 7)
    for i := range hashTable {
        hashTable[i] = ""
    }

    entries := []string{"Bob:123", "Amy:456", "Eve:789"}

    for _, entry := range entries {
        key := ""
        for i := 0; i < len(entry); i++ {
            if entry[i] == ':' {
                key = entry[:i]
                break
            }
        }
        index := hash(key)
        for {
            if hashTable[index] == "" {
                hashTable[index] = entry
                break
            }
            index = (index + 1) % 7
        }
    }

    // Print the hash table below
    // Your code here
}

Hint

Use a for loop with range over hashTable. Print empty if the string is empty.

Practice

1. What is the main idea behind linear probing in collision handling for hash tables?

easy

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

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

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

   When two keys hash to the same slot, a collision occurs and must be resolved.

2. Step 2: Explain linear probing method

   Linear probing resolves collisions by checking the next slot in order until an empty one is found.

3. Final Answer:

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

4. Quick Check:

   Linear probing = sequential slot check [OK]

Hint: Linear probing checks next slots 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 m when current index is i?

easy

A. (i + 1) % m

B. (i + 3) % m

C. (i - 1) % m

D. (i * 2) % m

Solution

1. Step 1: Understand linear probing index calculation

   Linear probing moves to the next slot by adding 1 to the current index.

2. Step 2: Apply modulo to wrap around

   To stay within table bounds, use modulo with table size: (i + 1) % m.

3. Final Answer:

   (i + 1) % m -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Using multiplication instead of addition
• Not using modulo causing out-of-bound errors
• Subtracting instead of adding for next slot

3. Consider a hash table of size 5 with linear probing. Insert keys 7, 12, 17 in that order using hash function h(k) = k % 5. What is the final state of the table?

medium

A. [7, 12, 17, null, null]

B. [7, null, null, 12, 17]

C. [7, 12, null, 17, null]

D. [null, null, 7, 12, 17]

Solution

1. Step 1: Calculate initial positions

   h(7) = 7 % 5 = 2, h(12) = 12 % 5 = 2, h(17) = 17 % 5 = 2.

2. Step 2: Insert keys using linear probing

   Insert 7 at index 2. For 12, index 2 is taken, so check index 3 (empty), insert 12. For 17, indexes 2 and 3 taken, check index 4 (empty), insert 17.

3. Final Answer:

   [null, null, 7, 12, 17] -> Option D

4. Quick Check:

   All keys placed sequentially after collision [OK]

Hint: Check hash, then next slots if occupied [OK]

Common Mistakes:

• Placing keys at wrong indices ignoring collisions
• Not applying modulo for hash function
• Assuming keys overwrite existing ones

4. The following Go code snippet tries to insert keys into a hash table using linear probing. What is the error?

func insert(table []int, key int) {
    m := len(table)
    index := key % m
    for table[index] != -1 {
        index = (index + 1) % m
    }
    table[index] = key
}

// Initial table: [-1, -1, -1, -1, -1]
// Insert keys: 3, 8, 13

medium

A. The loop condition should check for zero instead of -1.

B. The code does not handle the case when the table is full, causing an infinite loop.

C. The initial table should be initialized with zeros, not -1.

D. The modulo operation should be done with key, not m.

Solution

1. Step 1: Analyze loop condition and termination

   The loop continues while the slot is occupied (not -1). If the table is full, the loop never ends.

2. Step 2: Identify missing full table check

   There is no check to stop after probing all slots, so infinite loop can occur if table is full.

3. Final Answer:

   The code does not handle the case when the table is full, causing an infinite loop. -> Option B

4. Quick Check:

   Missing full table check causes infinite loop [OK]

Hint: Always check for full table to avoid infinite loops [OK]

Common Mistakes:

• Ignoring full table condition
• Confusing initial empty slot value
• Misusing modulo operation

5. You have a hash table of size 7 using linear probing. The hash function is h(k) = k % 7. You insert keys 10, 20, 15, 7, 32 in this order. After these insertions, you want to insert key 24. Which index will key 24 be placed at?

hard

A. Index 3

B. Index 4

C. Index 5

D. Index 6

Solution

1. Step 1: Insert keys and find their positions

   h(10)=3, insert at 3. h(20)=6, insert at 6. h(15)=1, insert at 1. h(7)=0, insert at 0. h(32)=4, insert at 4.

2. Step 2: Find position for key 24

   h(24)=3, but index 3 occupied by 10. Check next index 4 (occupied by 32), then 5 (empty), so insert at 5.

3. Final Answer:

   Index 5 -> Option C

4. Quick Check:

   Linear probe from 3 to 5 for key 24 [OK]

Hint: Probe next slots until empty found [OK]

Common Mistakes:

• Stopping at first collision without probing
• Not applying modulo for wrap-around
• Misplacing keys ignoring occupied slots