[LeetCode in Python] 347 (M) top k frequent elements top K high frequency elements

topic

https://leetcode-cn.com/problems/top-k-frequent-elements/

Given a non-empty array of integers, return the element with the highest k before its frequency.

Example 1:

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

Example 2:

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

Description:

You can assume that a given k is always reasonable, and 1 ≤ k ≤ the number of different elements in the array.
The time complexity of your algorithm must be better than O (n log n), where n is the size of the array.

Problem-solving ideas

• bucket sorting
• Scan the array first to get the digital frequency table
• Then construct a dictionary like {frequency: [number]}, which is called a bucket
• In the previous step, get the maximum frequencymax_f
• From max_fto 0reverse cycle, according to the meaning of the question, no need to consider abnormal conditions, the whole bucket is added to the result list
• Exit until the length of the result list meets k

Code

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        
        # - statistic frequency
        freq_dict = defaultdict(int)
        for n in nums:
            freq_dict[n] += 1
            
        # - assign buckets, format is {freq:[number]}
        bucket_dict = defaultdict(list)
        max_f = 0
        for n,f in freq_dict.items():
            bucket_dict[f].append(n)
            max_f = max(max_f, f)
        
        # - reverse traverse
        res = []
        for f in range(max_f,0,-1):
            if f in bucket_dict:
                res += bucket_dict[f]
                if len(res) >= k:
                    return res