LeetCode | 0347. Top K Frequent Elements first K-frequency elements [Python]

LeetCode 0347. Top K Frequent Elements [first K-frequency element Python Medium] [] [] bucket sort

Problem

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.

problem

Power button

Given a non-null integer array, the frequency of occurrence of which the first returns the k higher element.

Example 1:

输入: nums = [1,1,1,2,2,3], k = 2
输出: [1,2]

Example 2:

输入: nums = [1], k = 1
输出: [1]

Description:

• You can always assume a given rational k, and 1 ≤ k ≤ number of different elements in the array.
• Your time complexity of the algorithm must be better than O ( the n-the n-log ), the n- array sizes.

Thinking

Bucket sort

1. Frequency, the same element to element statistical occurrence count + 1.
2. Then bucket sort, the elements in the array, in accordance with the frequency of occurrence packets, i.e. the frequency of occurrence of the presence of the i-th element i of the tub.
3. Finally, before removing the reverse k elements from the bucket.

Time complexity : O (n)
space complexity : O (n)

Python code

class Solution(object):
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        # 统计元素的频率
        count = dict()
        for num in nums:
            count[num] = count.get(num, 0) + 1  # 返回字典 count 中 num 元素对应的值, 没有就赋值为 0
        
        # 桶排序
        bucket = [[] for i in range(len(nums) + 1)]
        for key, value in count.items():
            bucket[value].append(key)
        
        # 逆序取出前 k 个元素
        res = list()
        for i in range(len(nums), -1, -1):  # 最后一个 -1 表示逆序
            if bucket[i]:
                res.extend(bucket[i])  # 在列表末尾追加元素
            if len(res) >= k:  # 只要前 k 个
                break
        return res[:k]  # 输出第 k 个之前的元素（包括第 k 个元素)

Code address

GitHub link