feat(0347_top-k-frequent-elements): add py3 soln

0347_top-k-frequent-elements/README.md

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+Given an integer array `nums` and an integer `k`, return _the_ `k` _most frequent elements_. You may return the answer in **any order**.
+
+**Example 1:**
+
+    Input: nums = [1,1,1,2,2,3], k = 2
+    Output: [1,2]
+    
+
+**Example 2:**
+
+    Input: nums = [1], k = 1
+    Output: [1]
+    
+
+**Constraints:**
+
+*   `1 <= nums.length <= 105`
+*   `k` is in the range `[1, the number of unique elements in the array]`.
+*   It is **guaranteed** that the answer is **unique**.
+
+**Follow up:** Your algorithm's time complexity must be better than `O(n log n)`, where n is the array's size.

0347_top-k-frequent-elements/python3/smart.py

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+# Time complexity: O(nlogn)
+# https://github.com/python/cpython/blob/f52d987abfda25e50469c9b6fe1d19f72453d2de/Lib/collections/__init__.py#L608
+from collections import Counter
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        return [val for val, _ in Counter(nums).most_common(k)]

0347_top-k-frequent-elements/python3/solution.py

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+from heapq import heapify, heappop
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        c = Counter(nums)
+        
+        # Python heapq creates a min-heap but we want a max-heap built
+        # against counts. Tuples are comparable in Python by default so
+        # we can create a tuple of (count, element) to build the heap. To
+        # mimic a max-heap, we can just negate the counts.
+        heap = [(-count, val) for val, count in c.items()]
+        heapify(heap)
+        
+        result = [None] * k
+        
+        for i, _ in enumerate(result):
+            result[i] = heappop(heap)[1]
+        
+        return result