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TopKFrequentWords.java
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TopKFrequentWords.java
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package heap;
import java.util.*;
/**
* Created by gouthamvidyapradhan on 07/04/2018.
* 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: O(n log k). Calculate frequency and maintain a inverse priority queue of size k and add elements. Return
result by reversing the priority queue elements.
*/
public class TopKFrequentWords {
class Pair{
String word;
int freq;
Pair(String word, int freq){
this.word = word;
this.freq = freq;
}
}
/**
* Main method
* @param args
* @throws Exception
*/
public static void main(String[] args) throws Exception{
String[] words = {"i", "love", "leetcode", "i", "love", "coding"};
List<String> sorted = new TopKFrequentWords().topKFrequent(words, 2);
sorted.stream().forEach(System.out::println);
}
public List<String> topKFrequent(String[] words, int k) {
Map<String, Integer> map = new HashMap<>();
for(String w : words){
map.putIfAbsent(w, 0);
int freq = map.get(w);
map.put(w, freq + 1);
}
Queue<Pair> pq = new PriorityQueue<>((o1, o2) -> (o1.freq == o2.freq) ? o2.word.compareTo(o1.word) :
Integer.compare(o1.freq, o2.freq));
for(String w : map.keySet()){
int f = map.get(w);
pq.offer(new Pair(w, f));
if(pq.size() > k){
pq.poll();
}
}
List<String> result = new ArrayList<>();
while(!pq.isEmpty()){
result.add(pq.poll().word);
}
Collections.reverse(result);
return result;
}
}