0347: Top K Frequent Elements
Problem Statement
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]
Example 3:
Input: nums = [1,2,1,2,1,2,3,1,3,2], k = 2 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.
Code Solution
from collections import Counter
class Solution:
def topKFrequent(self, nums, k):
freq = Counter(nums)
buckets = [[] for _ in range(len(nums) + 1)]
for num, count in freq.items():
buckets[count].append(num)
res = []
for i in range(len(buckets) - 1, 0, -1):
for num in buckets[i]:
res.append(num)
if len(res) == k:
return res