Collision Handling Using Open Addressing Linear Probing in DSA Go - 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 fast insertions and searches happen as the table fills up.

How does the time to find or add an item change when more items are stored?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func linearProbingInsert(table []int, key int) {
  size := len(table)
  index := key % size
  for i := 0; i < size; i++ {
    probeIndex := (index + i) % size
    if table[probeIndex] == -1 {
      table[probeIndex] = key
      return
    }
  }
}
    

This code tries to insert a key into a hash table using linear probing to find the next empty slot if a collision occurs.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that checks each slot starting from the hashed index until it finds an empty slot.
• How many times: In the worst case, it can check every slot in the table once (up to n times, where n is table size).

How Execution Grows With Input

As the table size and number of stored keys grow, the number of checks needed to find an empty slot or a key grows too.

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 number of slots checked when collisions happen often.

Final Time Complexity

Time Complexity: O(n)

This means in the worst case, inserting or searching can take time proportional to the size of the table because it may check many slots one by one.

Common Mistake

[X] Wrong: "Insertion always takes constant time because hashing gives direct access."

[OK] Correct: When collisions happen, linear probing may check many slots, so time can grow with table size, not stay constant.

Interview Connect

Understanding how collision handling affects time helps you explain trade-offs in hash tables and shows you can analyze real data structure performance.

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 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