Collision Handling Using Open Addressing Linear Probing in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the hash index for a given key.

DSA Java

int hashIndex = key [1] tableSize;

Drag options to blanks, or click blank then click option'

A/

B+

C%

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using division (/) instead of modulo (%) causes incorrect indexing.

Using multiplication (*) or addition (+) changes the index unpredictably.

The modulo operator (%) is used to ensure the hash index stays within the table size.

2fill in blank

medium

Complete the code to find the next index in linear probing.

DSA Java

index = (index [1] 1) [2] tableSize;

Drag options to blanks, or click blank then click option'

A/

B-

C%

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction (-) moves backward, which is incorrect for linear probing.

Using division (/) instead of modulo (%) breaks the wrap-around logic.

In linear probing, we move to the next slot by adding 1 and then taking modulo to wrap around.

3fill in blank

hard

Fix the error in the insertion loop to handle collisions correctly.

DSA Java

while (table[index] != null && table[index].key != key) {
    index = (index [1] 1) % tableSize;
}

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction (-) causes backward movement, which is incorrect.

Using multiplication (*) or division (/) breaks the linear probing logic.

We add 1 to the index to check the next slot during collision handling.

4fill in blank

hard

Fill both blanks to complete the search method using linear probing.

DSA Java

int index = key [1] tableSize;
while (table[index] != null && table[index].key != key) {
    index = (index [2] 1) % tableSize;
}

Drag options to blanks, or click blank then click option'

A%

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using division (/) instead of modulo (%) for index calculation.

Using subtraction (-) instead of addition (+) in the probing step.

The initial index is calculated using modulo (%). Then, we add 1 and modulo again to handle collisions.

5fill in blank

hard

Fill both blanks to complete the insertion method with linear probing collision handling.

DSA Java

int index = key [1] tableSize;
while (table[index] != null && table[index].key != key) {
    index = (index [2] 1) % tableSize;
}
table[index] = new Entry(key, value);

Drag options to blanks, or click blank then click option'

A%

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication (*) or subtraction (-) instead of addition (+) in the probing step.

Not using modulo (%) to wrap around the table size.

Calculate initial index with modulo (%), then add 1 and modulo again to find the next free slot. Finally, insert the new entry at the found index.

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