Leetcode Problem 692 Top K Frequent Words.txt

692. Top K Frequent Words
Given a non-empty list of words, return the k most frequent elements.

Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.

Example 1:
Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
Output: ["i", "love"]
Explanation: "i" and "love" are the two most frequent words.
    Note that "i" comes before "love" due to a lower alphabetical order.
Example 2:
Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
Output: ["the", "is", "sunny", "day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
    with the number of occurrence being 4, 3, 2 and 1 respectively.
Note:
You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Input words contain only lowercase letters.
Follow up:
Try to solve it in O(n log k) time and O(n) extra space.


Hint to solve: Use a heap. Maintain only K elements. Write a custom comparator when inserting into the heap.

import heapq
class CustomSort:
    
    def __init__(self, count, word):
        self.count = count
        self.word = word
    
    def __lt__(self, other):
        if self.count == other.count:
            return self.word > other.word
        return self.count < other.count
    
    def __eq__(self,other):
        return self.count == other.count and self.word == other.word


class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:
        
        result = []
        myPQ = []
        
        wordFreq = collections.Counter(words)
        
        for word,count in wordFreq.items():
            heapq.heappush(myPQ, (CustomSort(count,word), word))
            
            if(len(myPQ) > k):
                heapq.heappop(myPQ)
        
        for _ in range(0,k):
            result.append(heapq.heappop(myPQ)[1])
            
        result.reverse()
        return result