Top K Frequent Words
Created: February 29, 2020 by [lek-tin]
Last updated: February 29, 2020
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.
Solution
from collections import Counter
import heapq
class Pair:
def __init__(self, word, count):
self.word = word
self.count = count
Pair.__lt__ = lambda x, y: x.count > y.count or (x.count == y.count and x.word < y.word)
class Solution:
def topKFrequent(self, words: List[str], k: int) -> List[str]:
countMap = Counter(words)
for word in words:
try:
countMap[word] += 1
except:
countMap[word] = 1
q = []
for word, count in countMap.items():
heapq.heappush(q, Pair(word, count))
res = []
while k > 0:
res.append(heapq.heappop(q).word)
k -= 1
return res