From 47af3a05a88e972ea395cbebd1fdcd05f6110cc3 Mon Sep 17 00:00:00 2001
From: Laurel S <ls60@protonmail.com>
Date: Tue, 19 Jul 2022 20:48:06 -0700
Subject: [PATCH 1/2] Passes exercises 1 and 2

---
 hash_practice/exercises.py | 39 ++++++++++++++++++++++++++++++++++----
 1 file changed, 35 insertions(+), 4 deletions(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..163d3cc 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -1,19 +1,50 @@
 
+from ntpath import join
+from sqlalchemy import true
+
+
 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: ?
     """
-    pass
+    word_dict = {}
+    for word in strings:
+        sorted_word = ''.join(sorted(word))
+        if sorted_word in word_dict:
+            word_dict[sorted_word].append(word)
+            print(f"{word} found in dict")
+        else:
+            word_dict[sorted_word] = [word]
+    result = []
+    for k, v in word_dict.items():
+        result.append(v)
+    return result
+
+words = ["eat", "tea", "tan", "ate", "nat", "bat"]
+
+grouped_anagrams(words)
 
 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 log n) (This is the result of using the list sort method)
+        Space Complexity: O(n)
     """
-    pass
+    if not nums or k < 1:
+        return []
+    freq_map = {}
+    for num in nums:
+        freq_map[num] = 1 + freq_map.get(num, 0)
+    freq_list = [(num, freq) for num, freq in freq_map.items()]
+    freq_list.sort(key = lambda x: x[1], reverse=True)
+    top_el = [freq_list[i][0] for i in range(k)]
+    return top_el
+
+numbers = [1, 1, 1, 2, 2, 3]
+k = 2
+# top_k_frequent_elements(numbers, k)
 
 
 def valid_sudoku(table):

From af1acd53922932a78a667af3900a4ba05fa80943 Mon Sep 17 00:00:00 2001
From: Laurel S <ls60@protonmail.com>
Date: Tue, 19 Jul 2022 20:57:48 -0700
Subject: [PATCH 2/2] Adds time complexity analysis

---
 hash_practice/exercises.py | 21 +++++++--------------
 1 file changed, 7 insertions(+), 14 deletions(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 163d3cc..5826773 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -6,30 +6,27 @@
 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 * m log m)
+        ^ Not sure on the time complexity. Here, n represents the number of words in the 
+        input string, while m represents the length of an individual word. 
+        Space Complexity: O(n)
     """
     word_dict = {}
     for word in strings:
         sorted_word = ''.join(sorted(word))
         if sorted_word in word_dict:
             word_dict[sorted_word].append(word)
-            print(f"{word} found in dict")
         else:
             word_dict[sorted_word] = [word]
-    result = []
-    for k, v in word_dict.items():
-        result.append(v)
+    result = [v for k,v in word_dict.items()]
     return result
 
-words = ["eat", "tea", "tan", "ate", "nat", "bat"]
-
-grouped_anagrams(words)
 
 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: O(n log n) (This is the result of using the list sort method)
+        Time Complexity: O(n log n) (This is the result of using the list sort method and
+        n is the number of elements in nums)
         Space Complexity: O(n)
     """
     if not nums or k < 1:
@@ -42,10 +39,6 @@ def top_k_frequent_elements(nums, k):
     top_el = [freq_list[i][0] for i in range(k)]
     return top_el
 
-numbers = [1, 1, 1, 2, 2, 3]
-k = 2
-# top_k_frequent_elements(numbers, k)
-
 
 def valid_sudoku(table):
     """ This method will return the true if the table is still