692. 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:
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 URL

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Solution

找到一个stirng数组中，出现次数前k多的string。使用hash map 和 priority queue。

Using a hash map to build the relationship from word to frequence. Then use a priority queue to sort these Map.Entry<> from less frequent to most frequent and ordered anti-dictionary. When pq.size() > k, poll(). Then the remaining words in qp is the reverse of answer. Add them to 0 index of result list.

Code

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        List<String> res = new ArrayList<>();
        Map<String, Integer> map = new HashMap<>();
        for (String word : words){
            if (map.containsKey(word)){
                map.put(word, map.get(word) + 1);
            }
            else{
                map.put(word, 1);
            }
        }
        
        PriorityQueue<Map.Entry<String, Integer>> pq = new PriorityQueue(new Comparator<Map.Entry<String, Integer>>(){
            @Override
            public int compare(Map.Entry<String, Integer> a, Map.Entry<String, Integer> b){
                if (a.getValue() == b.getValue()){
                    return b.getKey().compareTo(a.getKey());
                }
                else{
                    return a.getValue() - b.getValue();
                }
            }
        });
        
        for (Map.Entry<String, Integer> m : map.entrySet()){
            pq.offer(m);
            if (pq.size() > k){
                pq.poll();
            }
        }
        
        while(!pq.isEmpty()){
            res.add(0, pq.poll().getKey());
        }
        return res;
    }
}

Time Complexity: O(nlogk)
Space Complexity: O(n)

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Review

We could create a priority queue much easier when using -> function.

PriorityQueue<T> pq = new PriorityQueue<>(
	 (a,b) -> a.getValue()==b.getValue() ? b.getKey().compareTo(a.getKey()) : a.getValue()-b.getValue()
);

To iterate a map using for loop, we have to convert it to

map.entrySet();