Collision Handling Using Open Addressing Linear Probing in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Linear Probing Insertions

What is the state of the hash table after inserting keys 10, 22, 31, 4, 15 into an empty hash table of size 7 using linear probing?

DSA C#

int size = 7;
int[] hashTable = new int[size];
for (int i = 0; i < size; i++) hashTable[i] = -1;
int[] keys = {10, 22, 31, 4, 15};
foreach (int key in keys) {
    int index = key % size;
    while (hashTable[index] != -1) {
        index = (index + 1) % size;
    }
    hashTable[index] = key;
}
// Print hashTable

A [10, 22, 31, 4, 15, -1, -1]

B [-1, 22, 15, 10, 31, 4, -1]

C [10, -1, 22, 31, 4, 15, -1]

D [10, 22, -1, 31, 4, 15, -1]

💡 Hint

Remember to use key % size to find initial index and then probe linearly for empty slot.

Each key is inserted at index = key % 7. If occupied, move to next index until empty slot found.

10 % 7 = 3 → slot 3 empty → insert 10

22 % 7 = 1 → slot 1 empty → insert 22

31 % 7 = 3 → slot 3 occupied by 10 → move to 4 empty → insert 31

4 % 7 = 4 → slot 4 occupied by 31 → move to 5 empty → insert 4

15 % 7 = 1 → slot 1 occupied by 22 → move to 2 empty → insert 15

❓ Predict Output

intermediate

Result of Searching in Linear Probing Hash Table

Given a hash table of size 5 with keys inserted using linear probing: [12, 25, 36, -1, 20], what will be the index returned when searching for key 36?

DSA C#

int size = 5;
int[] hashTable = {12, 25, 36, -1, 20};
int key = 36;
int index = key % size;
int startIndex = index;
while (hashTable[index] != key) {
    index = (index + 1) % size;
    if (index == startIndex || hashTable[index] == -1) {
        index = -1; // not found
        break;
    }
}
// index is the result

A 1

B 3

C 2

D -1

💡 Hint

Use key % size to find start index and probe until key found or empty slot.

36 % 5 = 1, start searching at index 1.

Index 1 has 25 (not 36), move to 2.

Index 2 has 36, found key at index 2.

🔧 Debug

advanced

Identify the Bug in Linear Probing Insert Method

What error will occur when running this linear probing insertion code snippet for a hash table of size 5?

DSA C#

int size = 5;
int[] hashTable = {1,2,3,4,5};
int key = 10;
int index = key % size;
while (hashTable[index] != -1) {
    index = (index + 1);
}
hashTable[index] = key;

A IndexOutOfRangeException

B No error, inserts key correctly

C Infinite loop

D NullReferenceException

💡 Hint

Check how index is updated inside the loop.

The index increments without modulo operation, so it can exceed array bounds causing IndexOutOfRangeException.

🧠 Conceptual

advanced

Why Use Linear Probing in Open Addressing?

Which of the following best explains the main advantage of using linear probing for collision handling in open addressing?

A It is simple to implement and uses contiguous slots for better cache performance

B It guarantees no collisions will occur

C It reduces clustering by spreading keys randomly across the table

D It uses linked lists to handle collisions efficiently

💡 Hint

Think about how linear probing searches for next free slot.

Linear probing is simple and uses contiguous slots, which improves cache locality and performance.

However, it can cause clustering, so option A is incorrect.

It does not guarantee no collisions (B) and does not use linked lists (D).

❓ Predict Output

expert

Final Hash Table State After Multiple Collisions

Consider a hash table of size 10 using linear probing. Insert keys in this order: 20, 30, 40, 50, 60, 70, 80. What is the final state of the hash table array?

DSA C#

int size = 10;
int[] hashTable = new int[size];
for (int i = 0; i < size; i++) hashTable[i] = -1;
int[] keys = {20, 30, 40, 50, 60, 70, 80};
foreach (int key in keys) {
    int index = key % size;
    while (hashTable[index] != -1) {
        index = (index + 1) % size;
    }
    hashTable[index] = key;
}
// Print hashTable

A [ -1, -1, -1, -1, -1, 20, 30, 40, 50, 80 ]

B [ -1, -1, -1, -1, -1, 20, 30, 40, 50, 60 ]

C [ -1, -1, -1, -1, -1, 20, 30, 40, 50, 70 ]

D [20, 30, 40, 50, 60, 70, 80, -1, -1, -1]

💡 Hint

Calculate key % 10 and probe linearly for each key.

All keys hash to index 0 (key % 10 = 0), so they probe linearly from index 0:

• 20 → index 0
• 30 → index 1
• 40 → index 2
• 50 → index 3
• 60 → index 4
• 70 → index 5
• 80 → index 6

Final hash table: [20, 30, 40, 50, 60, 70, 80, -1, -1, -1]

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