Collision Handling Using Open Addressing Linear Probing in DSA Java - 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 the time to insert or find an item changes as the table grows.

We ask: How many steps does it take to find a free spot or the item we want?

Scenario Under Consideration

Analyze the time complexity of the following insertion method using linear probing.

public void insert(int key) {
  int index = hash(key);
  while (table[index] != null) {
    index = (index + 1) % table.length;
  }
  table[index] = key;
}
    

This code tries to insert a key into a hash table. If the spot is taken, it moves one step forward until it finds an empty spot.

Identify Repeating Operations

Look for loops or repeated checks that happen during insertion.

• Primary operation: Checking if a table slot is empty and moving to the next slot.
• How many times: In the worst case, it can check many slots until it finds an empty one.

How Execution Grows With Input

As the table fills up, the number of checks before finding an empty slot grows.

Input Size (n)	Approx. Operations
10 (mostly empty)	1-2 checks
100 (half full)	Up to 50 checks
1000 (almost full)	Close to 1000 checks

Pattern observation: The more full the table, the longer it takes to find a free spot, growing roughly linearly with the number of items.

Final Time Complexity

Time Complexity: O(n)

This means in the worst case, the time to insert or find an item grows directly with the number of items in the table.

Common Mistake

[X] Wrong: "Insertion always takes constant time because hashing is fast."

[OK] Correct: When many collisions happen, linear probing may check many slots, making insertion slower as the table fills.

Interview Connect

Understanding how collision handling affects time helps you explain trade-offs in hash tables 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 slot is occupied, check the next slot sequentially until an empty one 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 occurs.

D. Rehash all elements using a new hash function.

Solution

1. Step 1: Understand linear probing concept

   Linear probing means when a collision happens, we look at the next slot in order, moving forward one by one.

2. Step 2: Compare options

   If a slot is occupied, check the next slot sequentially until an empty one is found. describes this exact process. Options B, C, and D describe other collision methods or resizing, not linear probing.

3. Final Answer:

   If a slot is occupied, check the next slot sequentially until an empty one 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 resizes the table
• Assuming linear probing uses a new hash function

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 * 2) % m

D. (i - 1) % m

Solution

1. Step 1: Recall linear probing index calculation

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

2. Step 2: Check options

   Only (i + 1) % m uses (i + 1) % m, which correctly moves to the next slot circularly.

3. Final Answer:

   (i + 1) % m -> Option B

4. Quick Check:

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

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

Common Mistakes:

• Adding 2 instead of 1
• Multiplying index instead of adding
• Subtracting index causing negative values

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

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Placing all keys at initial hash index
• Not wrapping around the table
• Confusing index positions

4. Given the following code snippet for linear probing insertion, what is the bug?

int insert(int[] table, int key) {
  int m = table.length;
  int i = key % m;
  while (table[i] != 0) {
    i = (i + 1) % m;
  }
  table[i] = key;
  return i;
}

medium

A. It does not handle the case when the table is full, causing an infinite loop.

B. The hash function should be key / m instead of key % m.

C. It inserts the key at the wrong index without probing.

D. It uses 0 to check empty slots, which is invalid for all keys.

Solution

1. Step 1: Analyze the while loop condition

   The loop continues while the slot is occupied (not zero). If the table is full, it loops forever because no empty slot is found.

2. Step 2: Check other options

   The hash function should be key / m instead of key % m. is wrong because modulo is correct for hashing. It inserts the key at the wrong index without probing. is wrong because probing is done. It uses 0 to check empty slots, which is invalid for all keys. assumes 0 is invalid, but 0 can be used as empty marker if keys are positive.

3. Final Answer:

   It does not handle the case when the table is full, causing an infinite loop. -> Option A

4. Quick Check:

   Full table causes infinite loop [OK]

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

Common Mistakes:

• Ignoring full table condition
• Misunderstanding hash function usage
• Assuming 0 cannot be empty slot marker

5. You have a hash table of size 7 using linear probing. The hash function is h(k) = k % 7. Insert keys 10, 24, 31, 17, 3, 7 in order. After all insertions, which index will key 7 be stored at?

hard

A. 3

B. 4

C. 1

D. 6

Solution

1. Step 1: Calculate initial hashes and insert keys

   h(10)=3, insert at 3.
   h(24)=3, slot 3 occupied, try 4.
   h(31)=3, slots 3 and 4 occupied, try 5.
   h(17)=3, slots 3,4,5 occupied, try 6.
   h(3)=3, slots 3,4,5,6 occupied, try 0.
   h(7)=0, slot 0 occupied, try 1.

2. Step 2: Check slot 1 for key 7

   Slot 1 is empty, so key 7 is stored at index 1.

3. Final Answer:

   1 -> Option C

4. Quick Check:

   Linear probing finds next free slot [OK]

Hint: Probe next slots until empty to find key position [OK]

Common Mistakes:

• Stopping at initial hash index
• Not checking all occupied slots
• Miscounting index after wrap-around