From 4413729f6e8582aa1cdd24dd363be86b70f6db11 Mon Sep 17 00:00:00 2001
From: Hannah Lumapas <lumapas.h@gmail.com>
Date: Mon, 17 Jan 2022 16:05:59 -0500
Subject: [PATCH] passes required tests

---
 hash_practice/exercises.py | 37 ++++++++++++++++++++++++++++++-------
 1 file changed, 30 insertions(+), 7 deletions(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..f400bbd 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -2,19 +2,43 @@
 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(n)
+        Space Complexity: O(n)
     """
-    pass
+    anagrams_hash = {}
+
+    for word in strings: 
+        sorted_word = "".join(sorted(word)) 
+        if sorted_word in anagrams_hash: 
+            anagrams_hash[sorted_word].append(word)
+        else: 
+            anagrams_hash[sorted_word] = [word]
+
+    return list(anagrams_hash.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: O(n)
+        Space Complexity: O(n)
     """
-    pass
+    frequency_hash = {}
+    if nums == []:
+        return nums
+
+    for num in nums:
+        if num in frequency_hash:
+            frequency_hash[num] += 1
+        else:
+            frequency_hash[num] = 1
 
+    return_list = []
+    for i in range(k):
+        highest_value = max(frequency_hash, key=frequency_hash.get)
+        return_list.append(highest_value)
+        frequency_hash.pop(highest_value)
+
+    return return_list
 
 def valid_sudoku(table):
     """ This method will return the true if the table is still
@@ -26,4 +50,3 @@ def valid_sudoku(table):
         Space Complexity: ?
     """
     pass
-