Top K Frequent Elements Using Heap in DSA Go - Time & Space Complexity

Time Complexity: Top K Frequent Elements Using Heap

O(n + m log k)

Understanding Time Complexity

We want to know how the time needed grows when finding the top K frequent elements using a heap.

How does the work change as the input list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func topKFrequent(nums []int, k int) []int {
    freqMap := make(map[int]int)
    for _, num := range nums {
        freqMap[num]++
    }

    h := &IntHeap{}
    heap.Init(h)
    for num, freq := range freqMap {
        heap.Push(h, Pair{num, freq})
        if h.Len() > k {
            heap.Pop(h)
        }
    }

    res := make([]int, 0, k)
    for h.Len() > 0 {
        res = append(res, heap.Pop(h).(Pair).num)
    }
    return res
}

// Pair and IntHeap types omitted for brevity
    

This code counts frequencies, then uses a min-heap to keep only the top K frequent elements.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop over all numbers to count frequencies, then loop over unique numbers to push/pop in heap.
• How many times: Counting loop runs n times (n = length of input), heap operations run m times (m = number of unique elements).

How Execution Grows With Input

As input size n grows, counting frequencies takes about n steps. Then, heap operations depend on m, the unique count.

Input Size (n)	Approx. Operations
10	~10 counting + ~10 heap ops
100	~100 counting + ~100 heap ops
1000	~1000 counting + ~1000 heap ops

Pattern observation: Counting grows linearly with n, heap operations grow with unique elements m, usually less than n.

Final Time Complexity

Time Complexity: O(n + m log k)

This means the time grows linearly with input size n for counting, and heap operations depend on m and k, keeping it efficient.

Common Mistake

[X] Wrong: "Heap operations run n times, so time is O(n log n)."

[OK] Correct: Heap size is limited to k, so each push/pop is O(log k), not O(log n), making it faster.

Interview Connect

Understanding how to use heaps to keep top K items efficiently is a valuable skill that shows you can handle large data smartly.

Self-Check

"What if we used a max-heap for all elements and then popped k times? How would the time complexity change?"

Practice

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

easy

A. To efficiently find and keep track of the top K frequent elements

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 role of heap in frequency tracking

   A heap helps keep track of elements based on their frequency efficiently, especially when we want only the top K frequent elements.

2. Step 2: Compare with other options

   Sorting all elements or removing duplicates is not the main goal here; the heap is used for efficient frequency comparison and retrieval.

3. Final Answer:

   To efficiently find and keep track of the top K frequent elements -> Option A

4. Quick Check:

   Heap usage = Efficient top K tracking [OK]

Hint: Heap keeps top K frequent elements efficiently [OK]

Common Mistakes:

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

2. Which of the following Go code snippets correctly initializes a min-heap to store pairs of (frequency, element) for the top K frequent elements problem?

easy

A. heap := new([]int); heap.Init(heap)

B. heap := make([]int, 0); heap.Push(heap, pair)

C. heap := &IntHeap{}; heap.Init(heap)

D. var h IntHeap; heap.Init(&h)

Solution

1. Step 1: Understand Go heap initialization

   In Go, a heap is usually implemented by defining a type that implements heap.Interface, then initializing it with heap.Init(&heap).

2. Step 2: Analyze options

   var h IntHeap; heap.Init(&h) correctly declares a variable of type IntHeap and initializes it by passing its address to heap.Init. Other options misuse pointers or heap methods.

3. Final Answer:

   var h IntHeap; heap.Init(&h) -> Option D

4. Quick Check:

   Correct heap init = var h IntHeap; heap.Init(&h) [OK]

Hint: Use heap.Init with pointer to heap variable [OK]

Common Mistakes:

• Calling heap.Init on non-pointer
• Using make for heap without type
• Incorrect pointer usage with new([]int)

3. What will be the output of the following Go code snippet that finds top 2 frequent elements?

nums := []int{1,1,1,2,2,3}
result := topKFrequent(nums, 2)
fmt.Println(result)

Assuming topKFrequent is correctly implemented using a min-heap.

medium

A. [1, 3]

B. [2, 3]

C. [1, 2]

D. [3, 2]

Solution

1. Step 1: Count frequencies of elements

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

2. Step 2: Extract top 2 frequent elements

   The top 2 frequent elements are 1 (3 times) and 2 (2 times). So the result is [1, 2].

3. Final Answer:

   [1, 2] -> Option C

4. Quick Check:

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

Hint: Count frequencies, pick top K by frequency [OK]

Common Mistakes:

• Confusing element order in output
• Including less frequent elements
• Ignoring frequency counts

4. The following Go code snippet is intended to find the top 3 frequent elements but has a bug. What is the bug?

func topKFrequent(nums []int, k int) []int {
    freqMap := make(map[int]int)
    for _, num := range nums {
        freqMap[num]++
    }
    
    h := &IntHeap{}
    heap.Init(h)
    
    for num, freq := range freqMap {
        heap.Push(h, Pair{freq, num})
        if h.Len() > k {
            heap.Pop(h)
        }
    }
    
    res := make([]int, 0, k)
    for h.Len() > 0 {
        p := heap.Pop(h).(Pair)
        res = append(res, p.num)
    }
    return res
}

medium

A. Result list is in ascending frequency order, needs reversing

B. Heap is not initialized before use

C. Frequency map is not counting correctly

D. Heap size check should be < k, not > k

Solution

1. Step 1: Understand heap usage and result order

   The min-heap keeps smallest frequency on top, so popping elements gives ascending frequency order.

2. Step 2: Identify output order issue

   Appending popped elements directly results in ascending order, but top K frequent elements should be in descending order of frequency.

3. Final Answer:

   Result list is in ascending frequency order, needs reversing -> Option A

4. Quick Check:

   Output order issue = Result list is in ascending frequency order, needs reversing [OK]

Hint: Pop from min-heap gives ascending order; reverse result [OK]

Common Mistakes:

• Ignoring output order of popped elements
• Not initializing heap properly
• Incorrect frequency counting

5. Given a large list of integers where many elements have the same frequency, how should the Top K Frequent Elements Using Heap algorithm be adapted to return the top K elements sorted by frequency and then by element value ascending when frequencies tie?

hard

A. Use a map only without heap to track frequencies

B. Use a heap with custom comparison: first by frequency, then by element value ascending

C. Use a max-heap with frequency only, ignoring element value

D. Sort the entire list first, then pick first K elements

Solution

1. Step 1: Understand tie-breaking requirement

   When frequencies tie, elements should be ordered by their value ascending to maintain consistent order.

2. Step 2: Adapt heap comparison logic

   Implement a custom comparison in the heap that compares frequency first, and if equal, compares element values ascending.

3. Final Answer:

   Use a heap with custom comparison: first by frequency, then by element value ascending -> Option B

4. Quick Check:

   Custom heap comparison handles ties correctly [OK]

Hint: Add tie-breaker in heap comparison for stable ordering [OK]

Common Mistakes:

• Ignoring tie-breaker in heap comparison
• Sorting entire list unnecessarily
• Using max-heap without tie-break logic