347.Top K Frequent Elements
===

###### tags: `Medium`,`Array`,`Hash Table`,`Heap`,`Divide and Conquer`,`Sorting`

[347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)

### 題目描述

Given an integer array `nums` and an integer `k`, return *the* `k` *most frequent elements*. You may return the answer in **any order**.

### 範例

**Example 1:**

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

**Example 2:**

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

**Constraints**:

* 1 <= `nums.length` <= 10^5^
* -10^4^ <= `nums[i]` <= 10^4^
* `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.

### 解答
#### Javascript

function topKfrequent(nums, k) {
  const map = new Map();
  for (const num of nums) {
    map.set(num, map.get(num) + 1 || 1);
  }

  // sort
  const sorted = [...map.entries()].sort((a, b) => b[1] - a[1]);
  const result = [];
  for (let i = 0; i < k; i++) {
    result.push(sorted[i][0]);
  }

  return result;
}

> [name=Marsgoat][time=May 22, 2023]

#### TypeScript

function topKFrequent(nums: number[], k: number): number[] {
  const map = new Map<number, number>();

  nums.forEach((num) => map.set(num, (map.get(num) || 0) + 1));

  const sortedArr = [...map.entries()].sort((a, b) => b[1] - a[1]);
  
  // 這樣慢一點 但是可以一行
  return sortedArr.slice(0, k).map((num) => num[0]);
}

> [name=Sheep][time=May 22, 2023]

#### Python

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        if len(nums) == k: return nums
        counter = Counter(nums)

        # Max heap
        heap = [(-freq, num) for num, freq in counter.items()]
        heapq.heapify(heap)
        
        ans = []
        for _ in range(k):
            ans.append(heapq.heappop(heap)[1])

        return ans

看了題解發現可以縮

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]: 
        if k == len(nums): return nums
        counter = Counter(nums)   

        return heapq.nlargest(k, counter.keys(), key=counter.get)

TC: $O(nlogk)$
SC: $O(n)$
> [name=Ron Chen][time=Mon, May 22, 2023]

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        return [n for n, _ in Counter(nums).most_common(k)]

> [name=Yen-Chi Chen][time=Mon, May 22, 2023]

### Reference

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