leetcode_gym/_problems/topFrequentKElements.md

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layout	title	last_solved	category	leetcode_url	status
page	Top Frequent K Elements	2020-07-17	heap	https://leetcode.com/problems/top-k-frequent-elements/solution/	Attempted

Problem

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.
It's guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
You can return the answer in any order.

Solution

Two ways to solve this. Both involve heapq library

{% highlight python %}

from collections import Counter import heapq

nums = [1,1,1,2,2,3]

def topKFrequent1(nums, k):

count = Counter(nums)

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

def topKFrequent2(nums, k):

c = Counter(nums)   # {1: 3, 2: 2, 3: 1}


c = [(-v,k) for k,v in c.items()]    # (v,k)  [(3, 1), (2, 2), (1, 3)]
                                     # (-v,k) [(-3, 1), (-2, 2), (-1, 3)]

heapq.heapify(c)

output = []
for i in range(k):
    item = heapq.heappop(c)
    output.append(item[1])

return output

print(topKFrequent2(nums, 2))

{% endhighlight %}

---

{% highlight python %}

def topKFrequent(self, nums: List[int], k: int) -> List[int]:

minHeap=[]
c = Counter(nums)

for n in c.values():
    
    heapq.heappush(minHeap, n)
    
    if len(minHeap)>k:
        heapq.heappop(minHeap)


out = []        
for k,v in c.items():
    if v in minHeap:
        out.append(k)

return out

{% endhighlight %}