feat: Top K Frequent Words

JavaScript/0692. Top K Frequent Words.js

@@ -0,0 +1,49 @@
+// 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.
+
+/**
+ * @param {string[]} words
+ * @param {number} k
+ * @return {string[]}
+ */
+
+/** 1) Sorting */
+// Time O(n log n)
+// Space O(n)
+const topKFrequent = (words, k) => {
+  if (words == null || words.length === 0) return null;
+
+  const map = {};
+  for (const w of words) {
+    if (map[w] == null) map[w] = 0;
+    map[w]++;
+  }
+
+  const compare = (w1, w2) => {
+    if (map[w1] !== map[w2]) return map[w2] - map[w1];
+    return w1.localeCompare(w2);
+  };
+  return Object.keys(map).sort(compare).slice(0, k);
+};
+
+/** 2) Priority Queue */
+// JavaScript is lack of priority queue, check Python version

Python/0692. Top K Frequent Words.py

@@ -0,0 +1,39 @@
+# 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.
+
+"""
+Priority Queue
+"""
+# https://leetcode.com/problems/top-k-frequent-words/solution/
+#
+# Time O(N + klogN): heapq.heapify operation and counting operations are O(N), and each of k heapq.heappop operations are O(logN).
+# Space O(N), the space used to store our count.
+import heapq
+from collections import Counter
+
+
+class Solution:
+    def topKFrequent(self, words: List[str], k: int) -> List[str]:
+        count = Counter(words)
+        q = [(-freq, word) for word, freq in count.items()]
+        heapq.heapify(q)
+        return [heapq.heappop(q)[1] for _ in range(k)]