692. Top K Frequent Words.md

692. Top K Frequent Words

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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.

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Solution

• mine

  • Java same as 347. Top K Frequent Elements

  Runtime: 5 ms, faster than 99.58%, Memory Usage: 46.8 MB, less than 5.36% of Java online submissions

  public List<String> topKFrequent(String[] words, int k) {
      Map<String, Integer> map = new HashMap<>();
      for (String word : words) {
          //words's item all is not the same
          map.put(word, map.getOrDefault(word, -1) + 1);
      }
      PriorityQueue<String>[] queueArray = new PriorityQueue[words.length];
      for (String key : map.keySet()) {
          int count = map.get(key);
          if (queueArray[count] == null) {
              queueArray[count] = new PriorityQueue<String>();
          }
          queueArray[count].add(key);
      }
      List<String> res = new ArrayList<>();
      for (int i = queueArray.length - 1; i >= 0 && k > 0; i--) {
          if (queueArray[i] != null) {
              int size = queueArray[i].size();
              if (k >= size) {
                  while (queueArray[i].size() > 0) {
                      res.add(queueArray[i].poll());
                  }
                  k -= size;
              } else {
                  while (k > 0) {
                      res.add(queueArray[i].poll());
                      k--;
                  }
                  break;
              }
          }
      }
      return res;
  }
  

  • the leetcode solution

    • Sorting: Runtime: 7 ms, faster than 57.39%, Memory Usage: 46.2 MB, less than 5.36% of Java online submissions

    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);
    }
    

    • Heap: Runtime: 8 ms, faster than 17.01%, Memory Usage: 46.6 MB, less than 5.36% of Java online submissions

    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;
    }
    

  • the most votes Runtime: 7 ms, faster than 57.39%, Memory Usage: 46 MB, less than 5.36% of Java online submissions

  public List<String> topKFrequent(String[] words, int k) {
  
      List<String> result = new LinkedList<>();
      Map<String, Integer> map = new HashMap<>();
      for (int i = 0; i < words.length; i++) {
          if (map.containsKey(words[i]))
              map.put(words[i], map.get(words[i]) + 1);
          else
              map.put(words[i], 1);
      }
  
      PriorityQueue<Map.Entry<String, Integer>> pq = new PriorityQueue<>(
              (a, b) -> a.getValue() == b.getValue() ? b.getKey().compareTo(a.getKey()) : a.getValue() - b.getValue()
      );
  
      for (Map.Entry<String, Integer> entry : map.entrySet()) {
          pq.offer(entry);
          if (pq.size() > k)
              pq.poll();
      }
  
      while (!pq.isEmpty())
          result.add(0, pq.poll().getKey());
  
      return result;
  }