Problem Description

You are given an integer array nums and an integer k. Your task is to find and return the k most frequent elements from the array.

The problem asks you to identify which elements appear most often in the array and return the top k elements based on their frequency. The elements in your answer can be returned in any order.

For example:

• If nums = [1,1,1,2,2,3] and k = 2, the element 1 appears 3 times, 2 appears 2 times, and 3 appears 1 time. So the two most frequent elements are [1,2].
• If nums = [1] and k = 1, the only element 1 appears once, so the answer is [1].

The key points to understand:

• You need to count how many times each unique element appears in the array
• You need to identify the k elements with the highest frequencies
• The order of elements in your output doesn't matter
• You're guaranteed that the answer is unique (there won't be ties where multiple valid answers exist)

Intuition

To find the most frequent elements, we need to know how often each element appears. This naturally leads us to think about counting frequencies first. A hash table (or dictionary) is perfect for this - we can traverse the array once and count occurrences of each element.

Once we have the frequency counts, we need to identify the top k elements. This is essentially a "top k" problem, which has several classic approaches:

1. We could sort all elements by their frequencies and take the first k elements. This works but might be inefficient if we're sorting all elements when we only need k of them.

2. We could use a heap (priority queue) to maintain only the top k elements. A min heap of size k is particularly clever here - we keep adding elements, and whenever the heap size exceeds k, we remove the element with the smallest frequency. This way, the heap always contains the k most frequent elements.

3. Since Python's Counter class has a built-in most_common(k) method that efficiently returns the k most frequent elements, we can leverage this for a clean solution.

The heap approach is especially efficient when k is much smaller than the number of unique elements, as we only maintain a heap of size k rather than sorting all elements. The time complexity becomes O(n log k) instead of O(n log n) for sorting.

The key insight is breaking the problem into two steps: first count frequencies, then select the top k. This separation makes the problem much more manageable and allows us to use appropriate data structures for each step.

Solution Approach

The solution uses a hash table combined with a min heap to efficiently find the top k frequent elements.

Step 1: Count Frequencies First, we traverse the array and use a hash table cnt to count the occurrence of each element. In Python, we can use Counter(nums) which automatically creates a dictionary where keys are the unique elements and values are their frequencies.

Step 2: Build and Maintain a Min Heap We iterate through the hash table and use a min heap to keep track of the top k frequent elements. The heap stores pairs of (frequency, element).

The process works as follows:

• For each element and its count in the hash table, we push the pair (count, element) into the min heap
• If the heap size exceeds k, we pop the smallest element (the one with minimum frequency)
• By maintaining the heap size at most k, we ensure that only the k most frequent elements remain

Step 3: Extract Results After processing all elements, the heap contains exactly the k most frequent elements. We pop all elements from the heap and add them to our result array.

Implementation Details:

• The min heap property ensures that the element with the smallest frequency is always at the root
• When we push a new element and the heap size exceeds k, popping removes the element with the smallest frequency among the current k+1 elements
• This guarantees that we keep the k elements with the highest frequencies

Time Complexity: O(n log k) where n is the length of the input array. We traverse the array once (O(n)) and perform heap operations for each unique element, where each operation takes O(log k).

Space Complexity: O(n) for the hash table to store frequencies, plus O(k) for the heap.

The provided solution code uses Python's Counter.most_common(k) method, which internally implements a similar efficient approach to find the k most frequent elements, returning them as a list of (element, count) tuples. The list comprehension [x for x, _ in cnt.most_common(k)] extracts just the elements, discarding the counts.

Example Walkthrough

Let's walk through the solution with nums = [1,1,1,2,2,3] and k = 2.

Step 1: Count Frequencies We traverse the array and count each element's occurrences:

• Element 1 appears 3 times
• Element 2 appears 2 times
• Element 3 appears 1 time

Our frequency map becomes: {1: 3, 2: 2, 3: 1}

Step 2: Build and Maintain Min Heap (size k=2)

We process each element-frequency pair:

1. Push (3, 1) to heap

   • Heap: [(3, 1)]
   • Size = 1 ≤ k, keep it

2. Push (2, 2) to heap

   • Heap: [(2, 2), (3, 1)] (min heap orders by frequency)
   • Size = 2 ≤ k, keep both

3. Push (1, 3) to heap

   • Heap temporarily: [(1, 3), (3, 1), (2, 2)]
   • Size = 3 > k, so pop minimum frequency
   • Pop (1, 3)
   • Heap: [(2, 2), (3, 1)]

Step 3: Extract Results The heap contains our k=2 most frequent elements:

• (3, 1) - element 1 with frequency 3
• (2, 2) - element 2 with frequency 2

Extracting just the elements gives us: [1, 2]

The key insight is that by maintaining a min heap of size k, we automatically filter out elements with lower frequencies. When element 3 (frequency 1) tried to enter, it was immediately removed because we already had k=2 elements with higher frequencies.

Solution Implementation

Time and Space Complexity

The time complexity of this code is O(n + m log m) where n is the length of the input array and m is the number of unique elements. In the worst case where all elements are unique, m = n, making it O(n log n). The space complexity is O(m) for storing the counter dictionary.

However, the reference answer suggests O(n log k) time complexity, which would be achievable if we used a min-heap of size k instead of the most_common(k) method. The Counter constructor takes O(n) time to count frequencies. The most_common(k) method internally sorts all unique elements by frequency, which takes O(m log m) time, then returns the top k elements.

The reference answer's space complexity of O(k) would also apply to a heap-based solution that maintains only k elements. The current implementation uses O(m) space for the Counter object which stores all unique elements and their frequencies, not just the top k.

Common Pitfalls

1. Assuming the Input Array is Sorted or Elements are Unique

A common mistake is assuming that the input array has some special properties. The array can contain duplicates (that's the whole point - counting frequencies), and elements can appear in any order. Don't try to optimize based on assumptions about the input structure.

2. Using Sorting on the Entire Frequency Map

A pitfall is sorting all elements by frequency and then taking the first k elements:

# Inefficient approach - sorts ALL unique elements
frequency_counter = Counter(nums)
sorted_items = sorted(frequency_counter.items(), key=lambda x: x[1], reverse=True)
return [item[0] for item in sorted_items[:k]]

This has O(n log n) time complexity where n is the number of unique elements. The heap-based approach or most_common(k) is more efficient at O(n log k).

3. Incorrect Heap Implementation

When manually implementing with a heap, a common error is using a max heap instead of a min heap, or forgetting to limit the heap size:

import heapq
# Wrong: This keeps ALL elements in the heap
heap = []
for num, freq in frequency_counter.items():
    heapq.heappush(heap, (freq, num))  # Missing the size limit check

Correct implementation:

import heapq
heap = []
for num, freq in frequency_counter.items():
    heapq.heappush(heap, (freq, num))
    if len(heap) > k:
        heapq.heappop(heap)  # Maintain heap size at most k

4. Returning Frequencies Instead of Elements

When using Counter.most_common(k), it returns tuples of (element, count). A mistake is returning the entire tuples or just the counts:

# Wrong: Returns tuples
return frequency_counter.most_common(k)  # Returns [(1, 3), (2, 2)]

# Wrong: Returns counts instead of elements  
return [count for _, count in frequency_counter.most_common(k)]  # Returns [3, 2]

5. Not Handling Edge Cases

While the problem guarantees valid input, in real scenarios you should consider:

• What if k equals the length of the array?
• What if k equals the number of unique elements?
• What if the array is empty? (though the problem constraints prevent this)

The Counter.most_common(k) method handles these cases gracefully, but manual implementations might fail if not careful.