347. Top K Frequent Elements

Medium


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

 

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

 

Constraints:

  • 1 <= nums.length <= 105
  • -104 <= nums[i] <= 104
  • k is in the range [1, the number of unique elements in the array].
  • It is guaranteed that the answer is unique.

 

Follow up: Your algorithm's time complexity must be better than O(n log n), where n is the array's size.





 class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:

        # c = Counter(nums)
        # a = [key for key, _ in c.most_common(k) ]
        # return a

#         items = {}
#         for num in nums:
#             items[num] = items.get(num, 0)+1

#         q = sorted(list(items.keys()), key=lambda x: -items[x])
#         return q[:k]

# using heap
        items = {}
        for num in nums:
            items[num] = items.get(num, 0)+1

        data = [(-v, k) for k, v in items.items()]

        heap = data
        heapq.heapify(heap)
        ans = []
        for item in range(k):
            x = heapq.heappop(heap)
            ans.append(x[1])

        return ans

Random Note


time.perf_counter() always returns the float value of time in seconds. while pref_counter_ns() always gives the integer value of time in nanoseconds.


t1_start = perf_counter()
t1_stop = perf_counter()
print("Elapsed time:", t1_stop, t1_start)