Collision Handling Using Open Addressing Linear Probing in DSA C++ - Time & Space Complexity

Time Complexity: Collision Handling Using Open Addressing Linear Probing

O(n)

Understanding Time Complexity

When we use open addressing with linear probing to handle collisions in a hash table, we want to know how long it takes to find or insert an item.

We ask: How does the time grow as the table fills up?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int linearProbingInsert(int hashTable[], int size, int key) {
  int index = key % size;
  int startIndex = index;
  while (hashTable[index] != -1) {
    index = (index + 1) % size;
    if (index == startIndex) return -1; // Table full
  }
  hashTable[index] = key;
  return index;
}
    

This code tries to insert a key into a hash table using linear probing to find an empty spot.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that checks each slot in the table until it finds an empty one.
• How many times: In the worst case, it can check every slot once, up to the size of the table.

How Execution Grows With Input

As the table gets fuller, the number of checks before finding an empty slot grows.

Input Size (n)	Approx. Operations
10	Up to 10 checks
100	Up to 100 checks
1000	Up to 1000 checks

Pattern observation: The number of operations can grow linearly with the table size in the worst case.

Final Time Complexity

Time Complexity: O(n)

This means in the worst case, the time to insert or search grows directly with the number of slots in the table.

Common Mistake

[X] Wrong: "Linear probing always takes constant time because it just checks one slot at a time."

[OK] Correct: When many slots are filled, it may need to check many slots in a row, making the time grow with the table size.

Interview Connect

Understanding how linear probing affects time helps you explain hash table performance clearly and shows you know how collisions impact speed.

Self-Check

"What if we changed linear probing to quadratic probing? How would the time complexity change?"

Practice

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

easy

A. Rehash all elements to a new table when a collision occurs.

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

Solution

1. Step 1: Understand linear probing concept

   Linear probing handles collisions by checking the next slot in order if the original slot is occupied.

2. Step 2: Compare with other methods

   Other options describe chaining or resizing, not linear probing.

3. Final Answer:

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

4. Quick Check:

   Linear probing = sequential slot check [OK]

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

Common Mistakes:

• Confusing linear probing with chaining
• Thinking collision means resizing immediately
• Assuming random slot probing

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 * 2) % m

B. (i + 1) % m

C. (i - 1) % m

D. (i + 3) % m

Solution

1. Step 1: Recall linear probing formula

   Linear probing moves to the next slot by adding 1 to the current index, wrapping around using modulo.

2. Step 2: Verify modulo usage

   Modulo with table size m ensures index stays within bounds.

3. Final Answer:

   (i + 1) % m -> Option B

4. Quick Check:

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

Hint: Next slot in linear probing is (i+1) mod table size [OK]

Common Mistakes:

• Using multiplication instead of addition
• Not using modulo to wrap index
• Subtracting instead of adding

3. Consider a hash table of size 7 using linear probing. The hash function is h(k) = k % 7. Insert keys 10, 20, 15 in order. What is the final state of the table (index: value)?

medium

A. 0: null, 1: 15, 2: null, 3: 10, 4: null, 5: null, 6: 20

B. 0: null, 1: null, 2: 15, 3: 10, 4: 20, 5: null, 6: null

C. 0: null, 1: null, 2: 10, 3: 20, 4: 15, 5: null, 6: null

D. 0: null, 1: null, 2: 10, 3: 15, 4: 20, 5: null, 6: null

Solution

1. Step 1: Compute hashes and insert 10

   h(10) = 10 % 7 = 3, index 3 empty, insert 10 at 3.

2. Step 2: Insert 20

   h(20) = 20 % 7 = 6, index 6 empty, insert 20 at 6.

3. Step 3: Insert 15

   h(15) = 15 % 7 = 1, index 1 empty, insert 15 at 1.

4. Final Answer:

   0: null, 1: 15, 2: null, 3: 10, 4: null, 5: null, 6: 20 -> Option A

5. Quick Check:

   Positions 1:15, 3:10, 6:20 [OK]

Hint: Calculate hash, insert keys, probe next if occupied [OK]

Common Mistakes:

• Miscomputing hash values
• Not applying modulo correctly
• Placing keys at wrong indices

4. The following code snippet attempts to insert keys into a hash table using linear probing. Identify the bug:

int hash(int key, int size) { return key % size; }

void insert(int table[], int size, int key) {
  int index = hash(key, size);
  while (table[index] != -1) {
    index = (index + 1) % size;
  }
  table[index] = key;
}

int main() {
  int size = 5;
  int table[5] = {-1, -1, -1, -1, -1};
  insert(table, size, 5);
  insert(table, size, 10);
  insert(table, size, 15);
  insert(table, size, 20);
  insert(table, size, 25);
  insert(table, size, 30);
  return 0;
}

medium

A. Infinite loop possible when table is full because no stop condition in while loop.

B. Hash function is incorrect; should use (key + 1) % size.

C. No bug, code works correctly for all insertions.

D. Array initialization with -1 is invalid in C++.

Solution

1. Step 1: Analyze insertion loop

   The while loop probes next index until an empty slot (-1) is found.

2. Step 2: Check for full table scenario

   If the table is full, the loop will cycle endlessly without a stop condition.

3. Step 3: Confirm other options

   Hash function is correct modulo. Array initialization with -1 is valid. So only infinite loop is the bug.

4. Final Answer:

   Infinite loop possible when table is full because no stop condition in while loop. -> Option A

5. Quick Check:

   Full table causes infinite loop in linear probing [OK]

Hint: Check if loop stops when table is full [OK]

Common Mistakes:

• Ignoring full table infinite loop
• Thinking hash function is wrong
• Assuming array init with -1 is invalid

5. You have a hash table of size 10 using linear probing. The table currently stores keys at indices: 2, 3, 4, 5, 6. You want to insert key 12 where h(k) = k % 10. If index 2 is occupied, what sequence of indices will be checked before insertion? What index will key 12 be inserted at if indices 2 to 6 are occupied but index 7 is free?

hard

A. Check indices 2, 4, 6, 8; insert at 8

B. Check indices 2, 3, 4; insert at 4

C. Check indices 2, 3, 4, 5, 6, 7; insert at 7

D. Check indices 2, 3, 4, 5, 6; insert at 2

Solution

1. Step 1: Calculate initial index for key 12

   h(12) = 12 % 10 = 2, so start checking at index 2.

2. Step 2: Apply linear probing sequence

   Since indices 2, 3, 4, 5, 6 are occupied, check next indices sequentially: 2, 3, 4, 5, 6, then 7.

3. Step 3: Find first free slot

   Index 7 is free, so insert key 12 at index 7.

4. Final Answer:

   Check indices 2, 3, 4, 5, 6, 7; insert at 7 -> Option C

5. Quick Check:

   Linear probe checks sequential slots until free [OK]

Hint: Probe sequentially until free slot found [OK]

Common Mistakes:

• Skipping indices in probing
• Inserting at first occupied slot
• Not wrapping index with modulo