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

We want to understand how long it takes to insert items when collisions happen in a hash table using linear probing.

How does the time to find a free spot 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) { // -1 means empty
        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 handle collisions.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that checks each slot until an empty one is found.
• How many times: In the worst case, it can check every slot once if the table is almost full.

How Execution Grows With Input

As the table fills, the number of checks grows because more slots are occupied.

Input Size (n)	Approx. Operations
10 (mostly empty)	1-2 checks
100 (half full)	5-10 checks
1000 (almost full)	up to 1000 checks

Pattern observation: The more full the table, the longer it takes to find an empty slot, growing roughly linearly with table size.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Insertion always takes constant time because we just compute the hash once."

[OK] Correct: When collisions happen, we may need to check many slots, so insertion time can grow with table fullness.

Interview Connect

Understanding how collision handling affects insertion time helps you explain trade-offs in hash table design clearly and confidently.

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

B. Use a linked list to store multiple elements at the same index.

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

D. Rehash all elements using a new hash function when a collision occurs.

Solution

1. Step 1: Understand linear probing concept

   Linear probing handles collisions by checking the next slot in the array sequentially when the desired slot is occupied.

2. Step 2: Compare options with definition

   If a collision occurs, check the next slot sequentially until an empty slot is found. correctly describes this sequential checking. Other options describe different collision handling methods.

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 means try next slot one by one [OK]

Common Mistakes:

• Confusing linear probing with chaining
• Thinking rehashing happens on every collision
• Assuming table size doubles automatically

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. next_index = (i + 3) % m

B. next_index = (i * 2) % m

C. next_index = (i - 1) % m

D. next_index = (i + 1) % m

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.

2. Step 2: Match formula with options

   next_index = (i + 1) % m matches this formula: (i + 1) % m. Other options use incorrect increments.

3. Final Answer:

   next_index = (i + 1) % m -> Option D

4. Quick Check:

   Linear probing next index = (i + 1) mod m [OK]

Hint: Linear probing adds 1 to index modulo table size [OK]

Common Mistakes:

• Using multiplication instead of addition
• 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(k) = k % 5. Insert keys 7, 12, 17 in order. What is the final state of the table?

medium

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

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

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

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

Solution

1. Step 1: Calculate initial positions using hash function

   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 occupied, so try index 3. Insert 12 at index 3. For 17, indices 2 and 3 are occupied, so insert at index 4.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Placing all keys at same index ignoring collisions
• Not wrapping around table size
• Misplacing keys at wrong indices

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

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;
}

medium

A. The index update should subtract 1 instead of adding 1.

B. The table is not initialized with -1 for empty slots.

C. The while loop condition should check for table[index] == -1.

D. The hash function should use multiplication instead of modulo.

Solution

1. Step 1: Analyze the insert function logic

   The function checks for empty slots by comparing with -1, so the table must be initialized with -1 to mark empty slots.

2. Step 2: Identify missing initialization

   If the table is not initialized with -1, the condition table[index] != -1 may fail incorrectly, causing wrong insertions or infinite loops.

3. Final Answer:

   The table is not initialized with -1 for empty slots. -> Option B

4. Quick Check:

   Empty slots must be marked -1 [OK]

Hint: Initialize table slots with -1 before insertion [OK]

Common Mistakes:

• Ignoring table initialization
• Wrong loop condition in while
• Incorrect hash function usage

5. You have a hash table of size 7 using linear probing. Insert keys 10, 20, 15, 7, 32 in order using hash function h(k) = k % 7. After all insertions, what is the content of the table?

hard

A. [7, 15, null, 10, 32, null, 20]

B. [null, 15, 32, 10, null, 20, 7]

C. [null, 15, 32, 10, 20, 7, null]

D. [null, 15, 32, 10, 20, null, 7]

Solution

1. Step 1: Calculate initial indices for each key

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

2. Step 2: Insert keys using linear probing

   Insert 10 at index 3. Insert 20 at index 6. Insert 15 at index 1. Insert 7 at index 0. Insert 32 at index 4 (empty). Final table: [7, 15, null, 10, 32, null, 20].

3. Step 3: Match with options

   [7, 15, null, 10, 32, null, 20] matches the correct order. Nulls at indices 2 and 5.

4. Final Answer:

   [7, 15, null, 10, 32, null, 20] -> Option A

5. Quick Check:

   Keys placed at their hash or next free slot [OK]

Hint: Calculate hash, insert or probe next free slot [OK]

Common Mistakes:

• Not probing when slot is occupied
• Misplacing keys at wrong indices
• Ignoring table size wrap-around