LC 347. Top K Frequent Elements (Python Reforma)t

Author: raul-sauco

leetcode/top-k-frequent-elements.py

@@ -1,18 +1,24 @@
+# 347. Top K Frequent Elements
+# 🟠 Medium
+#
 # https://leetcode.com/problems/top-k-frequent-elements/
+#
+# Tags: Array - Hash Table - Divide and Conquer - Sorting
+#  - Heap (Priority Queue) - Bucket Sort - Counting - Quickselect
 
 import timeit
 from collections import Counter
 from typing import List
 
 
-# Iterate over the elements in nums creating a dictionary with num => num_times_seen
-# Return the k most seen elements in the dictionary. For simplicity this solution uses Counter.
+# Use the built-in Python Counter to count the elements and its
+# `most_common(n)` method to get the k most frequent.
 #
 # Time complexity: O(n*log(n)) - Sorting the dictionary by value.
 # Space complexity: O(n) - The size of the dictionary.
 #
-# Runtime: 163 ms, faster than 48.84% of Python3 online submissions for Top K Frequent Elements.
-# Memory Usage: 18.6 MB, less than 91.62% of Python3 online submissions for Top K Frequent Elements.
+# Runtime 95 ms Beats 95.19%
+# Memory Usage 18.4 Beats 99.66%
 class Solution:
     def topKFrequent(self, nums: List[int], k: int) -> List[int]:
         return [x[0] for x in Counter(nums).most_common(k)]
@@ -21,25 +27,27 @@ def topKFrequent(self, nums: List[int], k: int) -> List[int]:
 def test():
     executors = [Solution]
     tests = [
+        [[1], 1, [1]],
         [[3, 0, 1, 0], 1, [0]],
+        [[1, 1, 1, 2, 2, 3], 2, [1, 2]],
         [[1, 1, 1, 2, 2, 2, 2, 3], 3, [2, 1, 3]],
         [[1, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 3], 3, [3, 2, 1]],
-        [[1, 1, 1, 2, 2, 3], 2, [1, 2]],
-        [[1], 1, [1]],
     ]
     for executor in executors:
         start = timeit.default_timer()
-        for _ in range(int(float("1"))):
+        for _ in range(1):
             for col, t in enumerate(tests):
                 sol = executor()
                 result = sol.topKFrequent(t[0], t[1])
                 exp = t[2]
-                assert (
-                    result.sort() == exp.sort()
-                ), f"\033[93m» {result} <> {exp}\033[91m for test {col} using \033[1m{executor.__name__}"
+                assert result == exp, (
+                    f"\033[93m» {result} <> {exp}\033[91m for"
+                    + f" test {col} using \033[1m{executor.__name__}"
+                )
         stop = timeit.default_timer()
         used = str(round(stop - start, 5))
-        res = "{0:20}{1:10}{2:10}".format(executor.__name__, used, "seconds")
+        cols = "{0:20}{1:10}{2:10}"
+        res = cols.format(executor.__name__, used, "seconds")
         print(f"\033[92m» {res}\033[0m")