completes all but extra credit

hash_practice/exercises.py

@@ -1,21 +1,77 @@
-
 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
+
+    anagram_result = {}
+
+    # iterate through each word in strings
+    for word in strings:
+        # sort anagrams into different lists
+        # use sorted to rearrange each word alphabetically
+        # eat, tea, ate --> aet, aet, aet
+        # nat --> ant
+        # bat --> abt 
+
+        # can I use this sorted method?
+        sort_anagram = ''.join(sorted(word))
+        if sort_anagram in anagram_result:
+            # if word is already an anagram of another word, append word to same list
+            anagram_result[sort_anagram].append(word)
+
+        else:
+            # add word to new list
+            # ! brackets need to be around word b/c we're adding it as a new list in hash table
+            anagram_result[sort_anagram] = [word]
+
+    return list(anagram_result.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: ?
     """
-    pass
 
+    #  return empty list if nums is empty
+    if not nums:
+        return []
+
+    # create an empty hash table. key = number, value = occurence in string 
+    # iterate through nums 
+    # if number is not currently in the table, add it to the table and update value to 1
+    # if number is in table, update value
+    nums_table = {}
+    for number in nums:
+        if number in nums_table:
+            nums_table[number] += 1
+        else:
+            nums_table[number] = 1
+
+
+    # create an empty array 
+    # iterate through hash table and append key to array based on value - in desc order
+    # delete key from original hash table so we don't see it again
+    nums_arr = []
+
+    # loop through K # of times
+    for n in range(k):
+        max_number = 0
+        frequency = 0
 
+        for num, count in nums_table.items():
+            if count > frequency:
+                frequency = count
+                max_number = num
+
+        nums_arr.append(max_number)
+        del nums_table[max_number]
+
+    return nums_arr
+
 def valid_sudoku(table):
     """ This method will return the true if the table is still
         a valid sudoku table.