Top K Frequent Words

Description

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:

1. You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
2. Input words contain only lowercase letters.

Follow up:

1. Try to solve it in O(n log k) time and O(n) extra space.

Solution(javascript)

class Heap {
  constructor(list, compare = (a, b) => a - b) {
    this.left = index => 2 * index + 1
    this.right = index => 2 * index + 2
    this.parent = index => Math.floor((index - 1) / 2)
    this.heapify = (index = 0) => {
      const { list } = this
      const leftIndex = this.left(index)
      const rightIndex = this.right(index)
      let maxIndex = index
      if (list[leftIndex] !== undefined
        && this.compare(list[maxIndex], list[leftIndex]) > 0) {
        maxIndex = leftIndex
      }
      if (list[rightIndex] !== undefined
        && this.compare(list[maxIndex], list[rightIndex]) > 0) {
        maxIndex = rightIndex
      }
      if (index !== maxIndex) {
        const temp = list[index]
        list[index] = list[maxIndex]
        list[maxIndex] = temp
        this.heapify(maxIndex)
      }
    }
    this.buildHeap = () => {
      for (let i = Math.floor(this.list.length / 2); i >= 0; i--) {
        this.heapify(i)
      }
      return this.list
    }
    this.extract = () => {
      const temp = this.list[0]
      this.list[0] = this.list[this.list.length - 1]
      this.list[this.list.length - 1] = temp
      const result = this.list.pop()
      this.heapify(0)
      return result
    }
    this.insert = (item) => {
      const { list } = this
      list.push(item)
      let index = list.length - 1
      let parentIndex = this.parent(index)
      while (list[parentIndex] !== undefined && this.compare(list[parentIndex], list[index]) > 0) {
        const temp = list[index]
        list[index] = list[parentIndex]
        list[parentIndex] = temp
        index = parentIndex
        parentIndex = this.parent(index)
      }
    }
    this.list = list
    this.compare = compare
    this.buildHeap()
  }
}

const topKFrequent = (words = [], k) => {
  const map = words.reduce((acc, word) => {
    acc[word] = (acc[word] || 0) + 1
    return acc
  }, {})
  const list = Object.keys(map)
  const maxHeap = new Heap(
    list,
    (wordA, wordB) => {
      if (map[wordA] === map[wordB]) {
        return wordA.localeCompare(wordB)
      }
      return map[wordB] - map[wordA]
    },
  )
  const result = []
  while (maxHeap.list.length > 0 && result.length < k) {
    result.push(maxHeap.extract())
  }
  return result
}