Top K Frequent Elements Using Heap in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a frequency map of elements.

DSA Rust

let mut freq_map = std::collections::HashMap::new();
for &num in nums.iter() {
    *freq_map.entry(num).or_insert([1]) += 1;
}

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

ANone

B0

Cnum

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 as initial count causes double counting.

We start counting from zero for each new element in the frequency map.

2fill in blank

medium

Complete the code to push elements with their frequency into a max-heap.

DSA Rust

let mut heap = std::collections::BinaryHeap::new();
for (&num, &count) in freq_map.iter() {
    heap.push(([1], num));
}

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Acount

B-count

Cnum

Dcount as i32

Attempts:

3 left

💡 Hint

Common Mistakes

Using -count simulates min-heap, popping least frequent first.

We use count directly because Rust's BinaryHeap is a max-heap, popping the largest frequencies first.

3fill in blank

hard

Fix the error in popping top k elements from the heap.

DSA Rust

let mut result = Vec::new();
for _ in 0..k {
    if let Some((_, num)) = heap.[1]() {
        result.push(num);
    }
}

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Atake

Bpeek

Cremove

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() causes infinite loop or repeated elements.

pop() removes and returns the top element from the heap.

4fill in blank

hard

Fill both blanks to build the frequency map and push into heap correctly.

DSA Rust

let mut freq_map = std::collections::HashMap::new();
for &num in nums.iter() {
    *freq_map.entry(num).or_insert([1]) += 1;
}
let mut heap = std::collections::BinaryHeap::new();
for (&num, &count) in freq_map.iter() {
    heap.push(([2], num));
}

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A0

Bcount

C-count

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting counts at 1 causes off-by-one errors.

Pushing -count results in min-heap behavior.

Initialize counts with 0 and push counts to leverage the max-heap.

5fill in blank

hard

Fill all three blanks to complete the function returning top k frequent elements.

DSA Rust

fn top_k_frequent(nums: Vec<i32>, k: usize) -> Vec<i32> {
    let mut freq_map = std::collections::HashMap::new();
    for &num in nums.iter() {
        *freq_map.entry(num).or_insert([1]) += 1;
    }
    let mut heap = std::collections::BinaryHeap::new();
    for (&num, &count) in freq_map.iter() {
        heap.push(([2], num));
    }
    let mut result = Vec::new();
    for _ in 0..k {
        if let Some((_, num)) = heap.[3]() {
            result.push(num);
        }
    }
    result
}

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

A0

Bcount

Cpop

D-count

Attempts:

3 left

💡 Hint

Common Mistakes

Using negative counts results in min-heap order.

Using peek() instead of pop() causes repeated elements.

Initialize counts at 0, push counts for max-heap order, and pop elements to get top k most frequent.

Practice

1. What is the main purpose of using a heap in the Top K Frequent Elements problem?

easy

A. To efficiently keep track of the top K elements by frequency

B. To sort all elements in ascending order

C. To store elements without any order

D. To remove duplicates from the list

Solution

1. Step 1: Understand the problem goal

   The goal is to find the K elements that appear most frequently in a list.

2. Step 2: Role of heap in this context

   A heap helps efficiently keep track of the top K frequent elements by maintaining a small structure that updates as we process frequencies.

3. Final Answer:

   To efficiently keep track of the top K elements by frequency -> Option A

4. Quick Check:

   Heap = Efficient top K tracking [OK]

Hint: Heap keeps top K elements efficiently by frequency [OK]

Common Mistakes:

• Thinking heap sorts all elements fully
• Confusing heap with simple list storage
• Assuming heap removes duplicates

2. Which Rust data structure is best suited to store frequencies of elements before using a heap?

easy

A. Vec<T>

B. HashMap

C. LinkedList<T>

D. BinaryHeap<T>

Solution

1. Step 1: Identify frequency storage needs

   We need to count how many times each element appears, so we need a key-value structure.

2. Step 2: Choose appropriate Rust structure

   HashMap stores elements as keys and their counts as values efficiently.

3. Final Answer:

   HashMap<T, usize> -> Option B

4. Quick Check:

   Frequency storage = HashMap [OK]

Hint: Use HashMap to count frequencies before heap [OK]

Common Mistakes:

• Using Vec which is not key-value
• Using LinkedList which is inefficient here
• Confusing BinaryHeap with frequency map

3. What is the output of this Rust code snippet?

use std::collections::{BinaryHeap, HashMap};

fn top_k_frequent(nums: Vec<i32>, k: usize) -> Vec<i32> {
    let mut freq = HashMap::new();
    for num in nums {
        *freq.entry(num).or_insert(0) += 1;
    }
    let mut heap = BinaryHeap::new();
    for (&num, &count) in &freq {
        heap.push((count, num));
    }
    let mut result = Vec::new();
    for _ in 0..k {
        if let Some((_, num)) = heap.pop() {
            result.push(num);
        }
    }
    result
}

fn main() {
    let nums = vec![1,1,1,2,2,3];
    let k = 2;
    let top = top_k_frequent(nums, k);
    println!("{:?}", top);
}

medium

A. [2, 1]

B. [3, 2]

C. [1, 2]

D. [1, 3]

Solution

1. Step 1: Count frequencies

   1 appears 3 times, 2 appears 2 times, 3 appears 1 time.

2. Step 2: Push into max heap and pop top k

   Heap stores (count, num). Popping twice gives (3,1) then (2,2), so result is [1, 2].

3. Final Answer:

   [1, 2] -> Option C

4. Quick Check:

   Top 2 frequent = [1, 2] [OK]

Hint: Heap pops highest frequency first [OK]

Common Mistakes:

• Confusing order of output elements
• Ignoring frequency counts
• Assuming heap sorts by element value

4. Identify the bug in this Rust code for top K frequent elements:

use std::collections::{BinaryHeap, HashMap};

fn top_k_frequent(nums: Vec<i32>, k: usize) -> Vec<i32> {
    let mut freq = HashMap::new();
    for num in nums {
        freq.insert(num, freq.get(&num).unwrap_or(&0) + 1);
    }
    let mut heap = BinaryHeap::new();
    for (&num, &count) in &freq {
        heap.push((count, num));
    }
    let mut result = Vec::new();
    for _ in 0..k {
        if let Some((_, num)) = heap.pop() {
            result.push(num);
        }
    }
    result
}

medium

A. Using freq.insert overwrites counts incorrectly

B. BinaryHeap cannot store tuples

C. Loop range 0..k is invalid

D. Result vector is not returned

Solution

1. Step 1: Analyze frequency counting

   Using freq.insert(num, freq.get(&num).unwrap_or(&0) + 1) overwrites previous counts without incrementing properly.

2. Step 2: Correct frequency update method

   Should use *freq.entry(num).or_insert(0) += 1; to increment counts correctly.

3. Final Answer:

   Using freq.insert overwrites counts incorrectly -> Option A

4. Quick Check:

   Incorrect freq update = Bug [OK]

Hint: Use entry().or_insert() to increment counts safely [OK]

Common Mistakes:

• Using insert instead of entry for counting
• Assuming BinaryHeap can't store tuples
• Ignoring return of result vector

5. Given a large list with many elements having the same frequency, how can you modify the heap approach to ensure the top K frequent elements are returned in ascending order of their values when frequencies tie?

hard

A. Use a min-heap with (frequency, -element) pairs

B. Use a max-heap with (frequency, element) pairs and reverse result after extraction

C. Use a max-heap with (frequency, Reverse(element)) pairs

D. Use a max-heap with (frequency, element) pairs and sort result by element ascending after extraction

Solution

1. Step 1: Understand tie-breaking need

   When frequencies tie, elements should be ordered ascending by their value in final output.

2. Step 2: Heap stores by frequency only

   Heap orders by frequency, but does not guarantee order among ties, so sorting result after extraction is needed.

3. Step 3: Sort extracted top K by element ascending

   After popping top K from heap, sort the result vector by element ascending to meet tie-break condition.

4. Final Answer:

   Use a max-heap with (frequency, element) pairs and sort result by element ascending after extraction -> Option D

5. Quick Check:

   Sort after heap extraction for tie-break [OK]

Hint: Sort top K result by element ascending after heap extraction [OK]

Common Mistakes:

• Trying to encode tie-break in heap key incorrectly
• Using negative values causing confusion
• Not sorting result after extraction