From 569734d4455a1ebea316842dc40c4f5d8a9e4011 Mon Sep 17 00:00:00 2001
From: Sangeeth Sudheer <sangeeth96@users.noreply.github.com>
Date: Sat, 9 Apr 2022 23:15:20 +0530
Subject: [PATCH] feat(0347_top-k-frequent-elements): add py3 soln

---
 0347_top-k-frequent-elements/README.md        | 21 +++++++++++++++++++
 0347_top-k-frequent-elements/python3/smart.py |  7 +++++++
 .../python3/solution.py                       | 19 +++++++++++++++++
 3 files changed, 47 insertions(+)
 create mode 100644 0347_top-k-frequent-elements/README.md
 create mode 100644 0347_top-k-frequent-elements/python3/smart.py
 create mode 100644 0347_top-k-frequent-elements/python3/solution.py

diff --git a/0347_top-k-frequent-elements/README.md b/0347_top-k-frequent-elements/README.md
new file mode 100644
index 0000000..ce10dc7
--- /dev/null
+++ b/0347_top-k-frequent-elements/README.md
@@ -0,0 +1,21 @@
+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.
\ No newline at end of file
diff --git a/0347_top-k-frequent-elements/python3/smart.py b/0347_top-k-frequent-elements/python3/smart.py
new file mode 100644
index 0000000..d5c6d30
--- /dev/null
+++ b/0347_top-k-frequent-elements/python3/smart.py
@@ -0,0 +1,7 @@
+# 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)]
diff --git a/0347_top-k-frequent-elements/python3/solution.py b/0347_top-k-frequent-elements/python3/solution.py
new file mode 100644
index 0000000..edb66e2
--- /dev/null
+++ b/0347_top-k-frequent-elements/python3/solution.py
@@ -0,0 +1,19 @@
+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