Approach #1: Sorting [Accepted]

Intuition and Algorithm

Count the frequency of each word, and sort the words with a custom ordering relation that uses these frequencies. Then take the best k of them.

Python

class Solution(object):
    def topKFrequent(self, words, k):
        count = collections.Counter(words)
        candidates = count.keys()
        candidates.sort(key = lambda w: (-count[w], w))
        return candidates[:k]

Java

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> count = new HashMap();
        for (String word: words) {
            count.put(word, count.getOrDefault(word, 0) + 1);
        }
        List<String> candidates = new ArrayList(count.keySet());
        Collections.sort(candidates, (w1, w2) -> count.get(w1).equals(count.get(w2)) ?
                w1.compareTo(w2) : count.get(w2) - count.get(w1));

        return candidates.subList(0, k);

Complexity Analysis

  • Time Complexity: , where is the length of words. We count the frequency of each word in time, then we sort the given words in time.

  • Space Complexity: , the space used to store our candidates.


Approach #2: Heap [Accepted]

Intuition and Algorithm

Count the frequency of each word, then add it to heap that stores the best k candidates. Here, "best" is defined with our custom ordering relation, which puts the worst candidates at the top of the heap. At the end, we pop off the heap up to k times and reverse the result so that the best candidates are first.

In Python, we instead use heapq.heapify, which can turn a list into a heap in linear time, simplifying our work.

Java

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> count = new HashMap();
        for (String word: words) {
            count.put(word, count.getOrDefault(word, 0) + 1);
        }
        PriorityQueue<String> heap = new PriorityQueue<String>(
                (w1, w2) -> count.get(w1).equals(count.get(w2)) ?
                w2.compareTo(w1) : count.get(w1) - count.get(w2) );

        for (String word: count.keySet()) {
            heap.offer(word);
            if (heap.size() > k) heap.poll();
        }

        List<String> ans = new ArrayList();
        while (!heap.isEmpty()) ans.add(heap.poll());
        Collections.reverse(ans);
        return ans;
    }
}
class Solution(object):
    def topKFrequent(self, words, k):
        count = collections.Counter(words)
        heap = [(-freq, word) for word, freq in count.items()]
        heapq.heapify(heap)
        return [heapq.heappop(heap)[1] for _ in xrange(k)]

Complexity Analysis

  • Time Complexity: , where is the length of words. We count the frequency of each word in time, then we add words to the heap, each in time. Finally, we pop from the heap up to times. As , this is in total.

In Python, we improve this to : our heapq.heapify operation and counting operations are , and each of heapq.heappop operations are .

  • Space Complexity: , the space used to store our count.

Analysis written by: @awice.