Add solution for Top K Frequent Elements

Author: jamiebase

top-k-frequent-elements/gcount85.py

@@ -0,0 +1,35 @@
+"""
+# Intuition
+N is can be up to 100000, so time complexity should be O(N log N) at least.
+Calculate the frequency of its elements using dictionary, sort it, and list top K ones.
+
+# Approach
+1. for n in nums:
+    1-1. if n in map: map[n] += 1
+2. sort it by its value
+3. slice and return top K
+
+# Complexity
+Let:
+- N = len(nums)
+- M = number of unique elements in nums
+
+Time complexity:
+- O(N) + O(M log M) + O(K)
+- O(N log N) when all elements are unique.
+
+Space complexity:
+- O(M) + O(M) + O(K)
+- Overall: O(N) in the worst case.
+"""
+
+from collections import defaultdict
+
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        dic = defaultdict(int)
+        for n in nums:
+            dic[n] += 1
+        value_sorted = sorted(dic.items(), key=lambda x: -x[1])
+        return [i[0] for i in value_sorted[:k]]