<reponame>yshshadow/Leetcode
# Given a non-empty array of integers, return the k most frequent elements.
#
# Example 1:
#
# Input: nums = [1,1,1,2,2,3], k = 2
# Output: [1,2]
# Example 2:
#
# Input: nums = [1], k = 1
# Output: [1]
# Note:
#
# You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
# Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
import collections


class Solution:
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        kv = collections.defaultdict(lambda: 0)
        for num in nums:
            kv[num] += 1


# Given a non-empty array of integers, return the k most frequent elements.
#
# Example 1:
#
# Input: nums = [1,1,1,2,2,3], k = 2
# Output: [1,2]
# Example 2:
#
# Input: nums = [1], k = 1
# Output: [1]
# Note:
#
# You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
# Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
import collections


class Solution:
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        kv = collections.defaultdict(lambda: 0)
        for num in nums:
            kv[num] += 1
# Given a non-empty array of integers, return the k most frequent elements. # # Example 1: # # Input: nums = [1,1,1,2,2,3], k = 2 # Output: [1,2] # Example 2: # # Input: nums = [1], k = 1 # Output: [1] # Note: # # You may assume k is always valid, 1 ≤ k ≤ number of unique elements. # Your algorithm's time complexity must be better than O(n log n), where n is the array's size. :type nums: List[int]         :type k: int         :rtype: List[int]