Top K Frequent Elements Using Heap in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using a heap to find the top K frequent elements?

We use a heap to keep track of the K elements with the highest frequency. This helps us efficiently find the most frequent elements without sorting the entire list.

intermediate

In the context of top K frequent elements, what type of heap is commonly used and why?

A min-heap is commonly used because it keeps the smallest frequency element at the top. This way, when the heap size exceeds K, we remove the smallest frequency element to keep only the top K frequent elements.

beginner

How do you build the frequency map before using the heap?

You count how many times each element appears in the list and store this count in a map or dictionary where keys are elements and values are their frequencies.

intermediate

Why is a heap more efficient than sorting the entire list when finding top K frequent elements?

Sorting the entire list takes O(n log n) time, but using a heap to keep track of top K elements takes O(n log k), which is faster when K is much smaller than n.

beginner

What is the output format after finding the top K frequent elements using a heap?

The output is a list of the K elements with the highest frequency, usually in any order or sorted by frequency depending on the implementation.

What data structure is used to efficiently find the top K frequent elements?

A Heap

B Linked List

C Stack

D Queue

Heap is used because it efficiently keeps track of the top K elements based on frequency.

Which heap type is best for keeping track of the top K frequent elements?

A Min-heap

B Binary Search Tree

C Max-heap

D Fibonacci Heap

Min-heap keeps the smallest frequency at the top, so we can remove it when the heap size exceeds K.

What is the first step before using a heap to find top K frequent elements?

A Sort the array

B Reverse the array

C Build a frequency map

D Create a linked list

We first count the frequency of each element and store it in a map.

Why is using a heap more efficient than sorting the entire list for top K frequent elements?

A Heap sorts the list faster

B Heap uses less memory

C Heap is easier to implement

D Heap has better average case time complexity for this problem

Heap operations take O(n log k) time, which is faster than sorting O(n log n) when K is small.

What happens when the heap size exceeds K during processing?

A The largest frequency element is removed

B The smallest frequency element is removed

C The heap is cleared

D Nothing happens

The smallest frequency element is removed to keep only the top K frequent elements.

Explain step-by-step how to find the top K frequent elements using a heap.

Think about counting first, then using a heap to keep only K elements.

Why is a min-heap preferred over a max-heap for the top K frequent elements problem?

Consider what happens when you have more than K elements in the heap.

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