algoadvance

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Clarifying Questions

  1. Input Size: Is there a constraint on the size of the input array nums?
    • Typical constraints might be: 1 <= nums.length <= 10^4 and k such that 1 <= k <= nums.length.
  2. Value Range: Are there any constraints on the values contained within nums?
    • Typically: -10^4 <= nums[i] <= 10^4.
  3. Uniqueness of k: Can the k frequent elements include ties (i.e., multiple elements having the same frequency)?
    • Yes, typically we just need to return any k elements that are most frequent.

Strategy

To solve this problem efficiently, we can use the following steps:

  1. Count frequency of elements: Use a Counter from the collections module to tally the frequency of each element in the array.
  2. Use a heap to find the top k elements:
    • Python’s heapq can help us maintain a heap of size k to keep track of the most frequent elements.
    • Utilize a min-heap because it allows efficient ways to keep the k most frequent elements by frequency.

Code

Let’s implement the solution using these steps:

from collections import Counter
import heapq

def topKFrequent(nums, k):
    # Step 1: Count frequency of each element
    count = Counter(nums)
    
    # Step 2: Use a heap to extract the k most frequent elements
    # Python's heapq is a min-heap
    heap = []
    
    for num, freq in count.items():
        heapq.heappush(heap, (freq, num))  # Push (frequency, num) tuple
        if len(heap) > k:
            heapq.heappop(heap)  # Ensure the heap keeps only k elements
    
    # Extract the numbers from the heap
    result = [num for freq, num in heap]
    
    return result

# Example usage
print(topKFrequent([1,1,1,2,2,3], 2))  # Output: [1, 2]
print(topKFrequent([1], 1))  # Output: [1]

Time Complexity

  1. Counting frequencies: This takes O(n) time, where n is the number of elements in nums.
  2. Heap operations: For each element in the count, pushing and popping from the heap takes O(log k) time.
    • Given there are at most n elements, heap operations will take O(n log k) time.
  3. Overall Complexity: The total time complexity is O(n + n log k), which simplifies to O(n log k).

By efficiently using a frequency count and a min-heap, this approach ensures that we can find the top k frequent elements within a reasonable time frame even for large input sizes.

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