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+'''
+692. Top K Frequent Words
+Medium
+
+6104
+
+291
+
+Add to List
+
+Share
+Given an array of strings words and an integer k, return the k most frequent strings.
+
+Return the answer sorted by the frequency from highest to lowest. Sort the words with the same frequency by their lexicographical order.
+
+ 
+
+Example 1:
+
+Input: words = ["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: words = ["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.
+ 
+
+Constraints:
+
+1 <= words.length <= 500
+1 <= words[i].length <= 10
+words[i] consists of lowercase English letters.
+k is in the range [1, The number of unique words[i]]
+ 
+
+Follow-up: Could you solve it in O(n log(k)) time and O(n) extra space?
+'''
+
+#Solution in Python 3.
+
+class Solution(object):
+    def topKFrequent(self, words, k):
+        """
+        :type words: List[str]
+        :type k: int
+        :rtype: List[str]
+        """
+        d = collections.defaultdict(int)
+        for word in words:
+            d[str(word)] += 1
+
+        # alternative ways to do that
+        # d = collections.Counter(words)
+        return [i[0] for i in sorted(d.items(), key = lambda x : (-x[1], x[0]))[:k]] # top K
+        # return [w for w, v in sorted(collections.Counter(words).items(), key = lambda x: (-x[1], x[0])) [:k]]