Collision Handling Using Open Addressing Linear Probing in DSA Javascript - 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 insert or find items as the table fills up.

We ask: How does the time to find a spot or search grow as more items are added?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function linearProbingInsert(hashTable, key) {
  let index = key % hashTable.length;
  while (hashTable[index] !== null) {
    index = (index + 1) % hashTable.length;
  }
  hashTable[index] = key;
}
    

This code tries to insert a key into a hash table using linear probing to find the next free spot if a collision happens.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

As the hash table fills, the number of checks to find an empty slot grows.

Input Size (n)	Approx. Operations
10 (mostly empty)	1 to 2 checks
100 (half full)	several checks, maybe 5-10
1000 (almost full)	up to 1000 checks in worst case

Pattern observation: When the table is mostly empty, insertion is fast. As it fills, the time to find a free spot grows 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 search grows linearly with the number of items in the table.

Common Mistake

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

[OK] Correct: When collisions happen and the table is full or nearly full, linear probing may check many slots, making insertion slower.

Interview Connect

Understanding how collision handling affects performance shows you grasp real-world data structure behavior, a key skill in coding interviews and software design.

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 values.

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

D. Replace the existing value in the slot with the new value.

Solution

1. Step 1: Understand collision handling methods

   Linear probing handles collisions by searching the next available slot in a sequence.

2. Step 2: Identify linear probing behavior

   When a collision happens, linear probing checks the next slot one by one until it finds an empty slot.

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 collision means resizing immediately
• Replacing existing values instead of 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 + 3) % m

D. (i + 1) % m

Solution

1. Step 1: Recall linear probing index update

   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 size, use modulo operation with table size m.

3. Final Answer:

   (i + 1) % m -> Option D

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:

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

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 (index: value)?

medium

A. 0: null, 1: null, 2: 7, 3: 12, 4: 17

B. 0: null, 1: 7, 2: null, 3: 12, 4: 17

C. 0: 7, 1: 12, 2: 17, 3: null, 4: null

D. 0: null, 1: 7, 2: 12, 3: 17, 4: null

Solution

1. Step 1: Calculate initial positions using hash function

   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 3 (empty), insert at 3. Insert 17: index 2 occupied, 3 occupied, try 4 (empty), insert at 4.

3. Final Answer:

   0: null, 1: null, 2: 7, 3: 12, 4: 17 -> Option A

4. Quick Check:

   Keys placed at 2, 3, 4 with probing [OK]

Hint: All keys hash to 2, probe next slots for empty [OK]

Common Mistakes:

• Placing all keys at index 2 ignoring collisions
• Skipping slots incorrectly during probing
• Miscomputing hash values

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

const size = 7;
const table = new Array(size).fill(null);
function hash(key) {
  return key % size;
}
function insert(key) {
  let index = hash(key);
  while (table[index] !== null) {
    index = (index + 1) % size;
  }
  table[index] = key;
}
insert(10);
insert(17);
insert(24);
console.log(table);

medium

A. The hash function should use key % (size - 1) instead of key % size.

B. The table array should be initialized with zeros instead of null.

C. The insert function does not handle the case when the table is full, causing an infinite loop.

D. The index update in the while loop should subtract 1 instead of adding 1.

Solution

1. Step 1: Analyze insert function loop

   The while loop probes next slots until it finds an empty slot (null).

2. Step 2: Identify missing full table check

   If the table is full, the loop cycles endlessly without exit condition, causing infinite loop.

3. Final Answer:

   The insert function does not handle the case when the table is full, causing an infinite loop. -> Option C

4. Quick Check:

   Missing full table check causes infinite loop [OK]

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

Common Mistakes:

• Thinking modulo should be size-1
• Ignoring infinite loop possibility
• Confusing null with zero initialization

5. You have a hash table of size 10 using linear probing. The table currently stores keys at indices: 2, 3, 4, 5. You want to insert key 22. The hash function is h(k) = k % 10. Which index will key 22 be inserted into?

hard

A. 2

B. 6

C. 7

D. 5

Solution

1. Step 1: Calculate initial index for key 22

   22 % 10 = 2, so initial index is 2.

2. Step 2: Apply linear probing to find empty slot

   Indices 2, 3, 4, 5 are occupied. Next index is 6, which is free.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Start at 2, probe until free slot at 6 [OK]

Hint: Probe forward from hash index until empty slot found [OK]

Common Mistakes:

• Inserting at first hashed index ignoring occupancy
• Stopping probing too early
• Choosing wrong next index