Medium
Top K Frequent Elements
Given an integer array 'nums' and an integer 'k', return the k most frequent elements within the array. The test cases are generated such that the answer is always unique. You may return the output in any order. Example 1: Input: nums = [1,2,2,3,3,3], k = 2 Output: [2,3] Example 2: Input: nums = [7,7], k = 1 Output: [7]
Constraints
- 1 <= nums.length <= 10^4. - -1000 <= nums[i] <= 1000 - 1 <= k <= number of distinct elements in nums.
Could you try grouping numbers as you loop through them? Look up 'Bucket Sort' algorithm. You are not limited to only one data structure, you can use two at the same time.
*hover to un-blur*
# Solution from Neetcode class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: count = {} freq = [[] for i in range(len(nums) + 1)] for num in nums: count[num] = 1 + count.get(num, 0) for num, cnt in count.items(): freq[cnt].append(num) res = [] for i in range(len(freq) - 1, 0, -1): for num in freq[i]: res.append(num) if len(res) == k: return res