Top K Frequent Elements Using Heap in DSA Rust - Execution Trace

Concept Flow - Top K Frequent Elements Using Heap

Count frequencies

↓

Build min-heap of size k

↓

For each element frequency

↓

If heap size < k

↓

Push freq

↓

Add to heap

↓

Heap contains top k frequencies

↓

Extract elements from heap

↓

Return top k frequent elements

Count element frequencies, keep a min-heap of size k with top frequencies, then extract results.

Execution Sample

DSA Rust

let nums = vec![1,1,1,2,2,3];
let k = 2;
// Count frequencies
// Use min-heap to keep top k
// Extract results

Find top 2 frequent elements from the list using a min-heap.

Execution Table

Step	Operation	Frequency Map	Heap Content (freq, elem)	Visual Heap State
1	Count frequencies	{1:3, 2:2, 3:1}	[]	Empty heap
2	Push (3,1) to heap	{1:3, 2:2, 3:1}	[(3,1)]	Heap: (3,1)
3	Push (2,2) to heap	{1:3, 2:2, 3:1}	[(2,2), (3,1)]	Heap: (2,2) root, (3,1) child
4	Check (1,3) freq=1 < min heap freq=2?	{1:3, 2:2, 3:1}	[(2,2), (3,1)]	No change, heap unchanged
5	Extract elements from heap	{1:3, 2:2, 3:1}	[(2,2), (3,1)]	Heap with top k frequencies
6	Return elements [1,2]	{1:3, 2:2, 3:1}	[]	Result: 1 and 2 are top frequent

💡 All frequencies processed, heap size reached k, extraction done.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
Frequency Map	{}	{1:3, 2:2, 3:1}	{1:3, 2:2, 3:1}	{1:3, 2:2, 3:1}	{1:3, 2:2, 3:1}	{1:3, 2:2, 3:1}
Heap Content	[]	[]	[(3,1)]	[(2,2), (3,1)]	[(2,2), (3,1)]	[]
Result List	[]	[]	[]	[]	[]	[1,2]

Key Moments - 3 Insights

Why do we use a min-heap instead of a max-heap to keep top k frequent elements?

We use a min-heap of size k to efficiently keep only the top k frequencies. The smallest frequency in the heap is at the root, so when a new frequency is higher, we replace the smallest. This keeps the heap size fixed and avoids storing all frequencies.

What happens when the heap size is less than k during insertion?

When the heap size is less than k, we simply push the new frequency element into the heap without removing any. This is shown in steps 2 and 3 where heap grows until size k.

Why do we skip pushing frequencies smaller than the min in the heap once heap size is k?

Because the heap already contains the top k frequencies, any frequency smaller than the minimum in the heap cannot be in the top k. So we skip them to keep the heap size and content correct.

Visual Quiz - 3 Questions

Look at the execution table, what is the heap content after step 3?

A [(3,1)]

B [(2,2), (3,1)]

C [(1,3), (2,2)]

D []

💡 Hint

Check the 'Heap Content' column in row for step 3.

At which step does the heap stop changing because the frequency is smaller than the min heap frequency?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Operation' and 'Heap Content' columns for step 4.

If k was 3 instead of 2, what would be the heap content after processing all frequencies?

A [(3,1), (2,2), (1,3)]

B [(3,1), (2,2)]

C [(1,3), (2,2)]

D [(3,1)]

💡 Hint

With k=3, all frequencies fit in the heap as shown in the variable tracker.

Concept Snapshot

Top K Frequent Elements Using Heap:
- Count frequencies of elements.
- Use a min-heap of size k to keep top frequencies.
- Push frequencies until heap size k.
- For new freq > min heap root, pop min and push new.
- Extract elements from heap for result.
- Efficient for large input with limited k.

Full Transcript

This visualization shows how to find the top k frequent elements using a min-heap. First, we count how many times each element appears. Then, we build a min-heap that holds at most k elements with the highest frequencies. We add frequencies to the heap until it reaches size k. After that, for each new frequency, if it is bigger than the smallest frequency in the heap, we remove the smallest and add the new one. This way, the heap always contains the top k frequencies. Finally, we extract the elements from the heap to get the answer. The execution table tracks each step, showing the frequency map, heap content, and heap visual state. The variable tracker shows how the frequency map and heap change over time. Key moments clarify why a min-heap is used and how the heap size is managed. The quiz tests understanding of heap content at different steps and the effect of changing k.

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