From 74e0207bf8f468defac63df6713eb39fac8c7ab1 Mon Sep 17 00:00:00 2001
From: Mary Tian <mary.tian.ic@smartsheet.com>
Date: Wed, 27 Jul 2022 20:08:44 -0600
Subject: [PATCH] Updated first two exercises.

---
 hash_practice/exercises.py | 34 ++++++++++++++++++++++++++++------
 1 file changed, 28 insertions(+), 6 deletions(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..b1a49d6 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -2,18 +2,40 @@
 def grouped_anagrams(strings):
     """ This method will return an array of arrays.
         Each subarray will have strings which are anagrams of each other
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: O(nlogn)? Not sure on this
+        Space Complexity: 0(n)
     """
-    pass
+    anagrams_dict = {}
+
+    for word in strings:
+        sorted_string = ''.join(sorted(word))
+
+        if sorted_string in anagrams_dict:
+            anagrams_dict[sorted_string].append(word)
+        else:
+            anagrams_dict[sorted_string] = [word]
+
+    return list(anagrams_dict.values())
 
 def top_k_frequent_elements(nums, k):
     """ This method will return the k most common elements
         In the case of a tie it will select the first occuring element.
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: 0(n)
+        Space Complexity: 0(n)
     """
-    pass
+    nums_dict = {}
+
+    if not nums:
+        return []
+
+    for num in nums:
+        if num in nums_dict:
+            nums_dict[num] += 1
+        else:
+            nums_dict[num] = 1
+
+    most_common_list = sorted(nums_dict, key=nums_dict.get, reverse=True)
+    return most_common_list[:k]
 
 
 def valid_sudoku(table):