692. Top K Frequent Words

Problem

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.

Problem Summary

Given a non-empty list of words, return the top k words with the highest frequency. The returned answer should be sorted by word frequency from highest to lowest. If different words have the same frequency, sort them in alphabetical order.

Solution Ideas

This is a very straightforward problem. Maintain a max heap of length k, first sorting by frequency, and if the frequencies are the same, sorting by alphabetical order. Finally, output by popping the elements from the priority queue one by one.

Code

package leetcode

import "container/heap"

func topKFrequent(words []string, k int) []string {
	m := map[string]int{}
	for _, word := range words {
		m[word]++
	}
	pq := &PQ{}
	heap.Init(pq)
	for w, c := range m {
		heap.Push(pq, &wordCount{w, c})
		if pq.Len() > k {
			heap.Pop(pq)
		}
	}
	res := make([]string, k)
	for i := k - 1; i >= 0; i-- {
		wc := heap.Pop(pq).(*wordCount)
		res[i] = wc.word
	}
	return res
}

type wordCount struct {
	word string
	cnt  int
}

type PQ []*wordCount

func (pq PQ) Len() int      { return len(pq) }
func (pq PQ) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] }
func (pq PQ) Less(i, j int) bool {
	if pq[i].cnt == pq[j].cnt {
		return pq[i].word > pq[j].word
	}
	return pq[i].cnt < pq[j].cnt
}
func (pq *PQ) Push(x interface{}) {
	tmp := x.(*wordCount)
	*pq = append(*pq, tmp)
}
func (pq *PQ) Pop() interface{} {
	n := len(*pq)
	tmp := (*pq)[n-1]
	*pq = (*pq)[:n-1]
	return tmp
}