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DSA Rustprogramming~10 mins

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

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to calculate the next index using linear probing.

DSA Rust
fn linear_probe(index: usize, step: usize, size: usize) -> usize {
    (index + [1]) % size
}
Drag options to blanks, or click blank then click option'
Aindex
Bstep
Csize
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Using the current index instead of the step for increment.
Not using modulo operation to wrap around the table.
2fill in blank
medium

Complete the code to insert a key into the hash table using linear probing.

DSA Rust
fn insert(table: &mut [Option<i32>], key: i32) {
    let size = table.len();
    let mut index = (key as usize) % size;
    while table[[1]].is_some() {
        index = (index + 1) % size;
    }
    table[index] = Some(key);
}
Drag options to blanks, or click blank then click option'
Asize
Bkey
Cindex
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Using the key instead of index to access the table.
Not updating the index correctly in the loop.
3fill in blank
hard

Fix the error in the search function to correctly find a key using linear probing.

DSA Rust
fn search(table: &[Option<i32>], key: i32) -> Option<usize> {
    let size = table.len();
    let mut index = (key as usize) % size;
    let start = index;
    loop {
        match table[[1]] {
            Some(k) if k == key => return Some(index),
            None => return None,
            _ => {
                index = (index + 1) % size;
                if index == start {
                    return None;
                }
            }
        }
    }
}
Drag options to blanks, or click blank then click option'
Aindex
Bkey
Cstart
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Using the key as an index to access the table.
Not checking for the loop wrap-around condition.
4fill in blank
hard

Fill both blanks to complete the delete function that marks a slot as deleted using linear probing.

DSA Rust
fn delete(table: &mut [Option<i32>], key: i32) {
    let size = table.len();
    let mut index = (key as usize) % size;
    let start = index;
    loop {
        match table[[1]] {
            Some(k) if k == key => {
                table[index] = [2];
                return;
            },
            None => return,
            _ => {
                index = (index + 1) % size;
                if index == start {
                    return;
                }
            }
        }
    }
}
Drag options to blanks, or click blank then click option'
Aindex
BNone
CSome(0)
DSome(key)
Attempts:
3 left
💡 Hint
Common Mistakes
Setting the slot to None instead of Some(0).
Using the key as an index.
5fill in blank
hard

Fill all three blanks to complete the function that counts occupied slots in the hash table.

DSA Rust
fn count_occupied(table: &[Option<i32>]) -> usize {
    table.iter().filter(|&[1]| [2].is_some() && [3] != 0).count()
}
Drag options to blanks, or click blank then click option'
Aref slot
Bslot
Cslot.unwrap_or(0)
Dslot.unwrap()
Attempts:
3 left
💡 Hint
Common Mistakes
Not using the correct pattern '|&slot|' in the closure parameter.
Comparing the Option directly without unwrapping.

Practice

(1/5)
1. What is the main idea behind linear probing in collision handling for hash tables?
easy
A. Use a secondary hash function to find a new slot.
B. Use a linked list to store multiple values at the same index.
C. Double the size of the hash table when a collision happens.
D. If a collision occurs, check the next slot sequentially until an empty slot is found.

Solution

  1. Step 1: Understand collision handling methods

    Linear probing is a method where, upon collision, the algorithm checks the next slot in the array sequentially.
  2. Step 2: Compare with other methods

    Other methods include chaining (linked lists), resizing, or double hashing, but linear probing specifically uses sequential slots.
  3. Final Answer:

    If a collision occurs, check the next slot sequentially until an empty slot is found. -> Option D
  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 linear probing uses a second hash function
  • Assuming table size doubles on collision
2. Which of the following Rust code snippets correctly implements linear probing to find the next index after a collision in a hash table of size table_size?
easy
A. let next_index = (current_index + 1) % table_size;
B. let next_index = (current_index * 2) % table_size;
C. let next_index = (current_index - 1) % table_size;
D. let next_index = (current_index + 3) % table_size;

Solution

  1. Step 1: Understand linear probing index calculation

    Linear probing moves to the next slot by adding 1 to the current index and wrapping around using modulo.
  2. Step 2: Check each option

    let next_index = (current_index + 1) % table_size; adds 1 and uses modulo, which is correct. Others multiply or subtract, which are not standard linear probing steps.
  3. Final Answer:

    let next_index = (current_index + 1) % table_size; -> Option A
  4. Quick Check:

    Linear probing = increment index by 1 modulo table size [OK]
Hint: Linear probing increments index by 1 modulo table size [OK]
Common Mistakes:
  • Using multiplication instead of addition
  • Subtracting index without handling negative wrap
  • Skipping modulo operation causing out-of-bounds
3. Consider the following Rust code snippet for inserting keys into a hash table using linear probing. What will be the final state of the hash table array after inserting keys [5, 15, 25] with table size 10 and hash function key % 10?
let mut table = vec![None; 10];
let keys = vec![5, 15, 25];
for &key in &keys {
    let mut index = key % 10;
    while table[index].is_some() {
        index = (index + 1) % 10;
    }
    table[index] = Some(key);
}
println!("{:?}", table);
medium
A. [None, None, None, None, None, Some(5), Some(15), None, Some(25), None]
B. [None, None, None, None, None, Some(5), None, Some(15), Some(25), None]
C. [None, None, None, None, None, Some(5), Some(15), Some(25), None, None]
D. [Some(5), Some(15), Some(25), None, None, None, None, None, None, None]

Solution

  1. Step 1: Calculate initial indices for keys

    Keys 5, 15, 25 all hash to index 5 (5 % 10 = 5, 15 % 10 = 5, 25 % 10 = 5).
  2. Step 2: Insert keys using linear probing

    Insert 5 at index 5. For 15, index 5 is occupied, so move to 6. Insert 15 at 6. For 25, indices 5 and 6 are occupied, move to 7 and insert 25.
  3. Final Answer:

    [None, None, None, None, None, Some(5), Some(15), Some(25), None, None] -> Option C
  4. Quick Check:

    Linear probing places keys at 5, 6, 7 [OK]
Hint: Keys with same hash go to next free slot sequentially [OK]
Common Mistakes:
  • Placing all keys at index 5 ignoring collisions
  • Skipping modulo wrap-around
  • Inserting keys at wrong indices after collision
4. The following Rust code attempts to insert keys into a hash table using linear probing. However, it sometimes causes an infinite loop. What is the bug?
let mut table = vec![None; 5];
let keys = vec![1, 6, 11];
for &key in &keys {
    let mut index = key % 5;
    while table[index].is_some() {
        index = (index + 1) % 5;
    }
    table[index] = Some(key);
}
medium
A. The code does not check if the table is full, causing infinite loop when no empty slot exists.
B. The modulo operator is used incorrectly; it should be key % 6.
C. The loop should break after checking only once to avoid infinite loop.
D. The vector size is too large, causing index out of bounds.

Solution

  1. Step 1: Analyze the while loop condition

    The loop continues while the slot is occupied, moving to next index with wrap-around.
  2. Step 2: Identify missing termination condition

    If the table is full, the loop cycles endlessly because no empty slot is found and no exit condition exists.
  3. Final Answer:

    The code does not check if the table is full, causing infinite loop when no empty slot exists. -> Option A
  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
  • Changing modulo divisor incorrectly
  • Breaking loop too early causing wrong insertion
5. You have a hash table of size 7 using linear probing for collision handling. You insert keys [10, 20, 15, 7, 32] with hash function key % 7. After all insertions, what is the state of the table? Represent empty slots as null. Explain your answer step-by-step.
medium
A. [10, 20, 15, 7, 32, null, null]
B. [7, 15, null, 10, 32, null, 20]
C. [null, 15, 32, 7, 10, 20, null]
D. [null, 15, 32, null, 10, null, 7]

Solution

  1. Step 1: Calculate initial indices for each key

    10 % 7 = 3, 20 % 7 = 6, 15 % 7 = 1, 7 % 7 = 0, 32 % 7 = 4.
  2. Step 2: Insert keys using linear probing

    All hashed indices are empty, so place: 10 at 3, 20 at 6, 15 at 1, 7 at 0, 32 at 4.
  3. Step 3: Final table state

    Indices 2 and 5 remain null: [7, 15, null, 10, 32, null, 20].
  4. Final Answer:

    [7, 15, null, 10, 32, null, 20] -> Option B
  5. Quick Check:

    Insert keys at hashed index or next free slot [OK]
Hint: Calculate hash, insert or probe next free slot [OK]
Common Mistakes:
  • Not applying modulo correctly
  • Placing keys at wrong indices after collision
  • Ignoring empty slots in final output
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10 questions~5 mineasy hard