add top-k-frequent-elements solution

Author: jongwanra

top-k-frequent-elements/jongwanra.py

@@ -0,0 +1,55 @@
+"""
+[Problem]
+https://leetcode.com/problems/top-k-frequent-elements/
+
+[Brain Storming]
+가장 자주 사용되는 요소 k개를 반환한다.
+
+1<= nums.length <= 10^5
+O(n^2) time complexity일 경우 시간 초과 발생
+
+[Plan]
+1. nums를 순회하면서 Hash Table에 key:element value:count 저장한다.
+2. nums를 순회하면서 Heap에 (count, element)를 count를 기준으로 Max Heap 형태로 저장한다.
+3. Max Heap에서 k개 만큼 pop한다.
+
+[Complexity]
+N: nums.length, K: k
+heapq.heapify(heap): O(N)
+heapq.heappop(heap): log(N)
+
+Time: O(N +  (K * log(N)))
+Space: O(N + K)
+    * num_to_count_map: O(N)
+    * heap: O(N)
+    * answer: O(K)
+
+
+"""
+
+import heapq
+
+class Solution(object):
+    def topKFrequent(self, nums, k):
+        num_to_count_map = {}
+        for num in nums:
+            num_to_count_map[num] = num_to_count_map.get(num, 0) + 1
+
+        heap = []
+        for num, count in num_to_count_map.items():
+            heap.append((-count, num))
+        heapq.heapify(heap)
+
+        answer = []
+        for _ in range(k):
+            negative_count, frequent_num = heapq.heappop(heap) # log(N)
+            answer.append(frequent_num)
+        return answer
+
+solution = Solution()
+
+# Normal Case
+print(solution.topKFrequent([1,1,1,2,2,3], 2) == [1, 2])
+print(solution.topKFrequent([1], 1) == [1])
+
+