From c08140a2309a15f83072b50b8098d5607d0e8755 Mon Sep 17 00:00:00 2001
From: marielaxcruz <marielacrtzcruz@gmail.com>
Date: Mon, 17 Jan 2022 20:05:42 -0800
Subject: [PATCH] Pass required tests

---
 hash_practice/exercises.py            | 46 +++++++++++++++++++++++----
 tests/test_top_k_frequent_elements.py |  2 +-
 2 files changed, 40 insertions(+), 8 deletions(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..530ba4c 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -1,20 +1,52 @@
 
-def grouped_anagrams(strings):
+def grouped_anagrams(d1):
     """ This method will return an array of arrays.
         Each subarray will have strings which are anagrams of each other
         Time Complexity: ?
         Space Complexity: ?
     """
-    pass
+    # empty dictionary for anagrams together 
+    dict = {} 
+
+    # traversal 
+    for val in d1: 
+        # sorts list
+        key = ''.join(sorted(val)) 
+
+        if key in dict.keys(): 
+            dict[key].append(val) 
+        else: 
+            dict[key] = [] 
+            dict[key].append(val) 
+
+    # traverse dictionary and join keys together 
+    result = []
+    for key,value in dict.items(): 
+        # result = result + ' '.join(value) + ' '
+        result.append(value)
+    return result 
 
-def top_k_frequent_elements(nums, k):
+def top_k_frequent_elements(numbers, 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
-
+    freq_map = {}
+    if numbers == []:
+        return numbers
+    for i in numbers:
+        if i in freq_map:
+            freq_map[i] += 1
+        elif i not in freq_map:
+            freq_map[i] = 1
+    
+    new_list = []
+    for i in range(k):
+        highest_value = max(freq_map, key=freq_map.get)
+        new_list.append(highest_value)
+        freq_map.pop(highest_value)
+    return new_list
 
 def valid_sudoku(table):
     """ This method will return the true if the table is still
diff --git a/tests/test_top_k_frequent_elements.py b/tests/test_top_k_frequent_elements.py
index 9e653cf..a44ed2c 100644
--- a/tests/test_top_k_frequent_elements.py
+++ b/tests/test_top_k_frequent_elements.py
@@ -7,7 +7,7 @@ def test_it_works_with_example_1():
     k = 2
 
     # Act
-    answer = top_k_frequent_elements(nums=numbers, k=k)
+    answer = top_k_frequent_elements(numbers=numbers, k=k)
 
     # Assert
     answer.sort()