maple - ruiz

hash_practice/exercises.py

@@ -2,18 +2,36 @@
 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(1)
     """
-    pass
+    anagram_dict = {}
+
+    for string in strings: 
+        word_sorted = "".join(sorted(string))
+        if word_sorted not in anagram_dict: 
+            anagram_dict[word_sorted] = [string]
+        else: 
+            anagram_dict[word_sorted] = [string]
+
+    return list(anagram_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: O(n)
+        Space Complexity: O(1)
     """
-    pass
+    most_common = {}
+    for num in nums: 
+        if num in most_common: 
+
+            most_common[num] += 1 
+        else: 
+            most_common[num] = 1 
+    most_common = sorted(most_common.items(), key=lambda x: x[1], reverse=True)
+    return [x[0] for x in most_common[:k]]
+
 
 
 def valid_sudoku(table):
@@ -22,8 +40,22 @@ def valid_sudoku(table):
         Each element can either be a ".", or a digit 1-9
         The same digit cannot appear twice or more in the same 
         row, column or 3x3 subgrid
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: O(n2)
+        Space Complexity: O(1)
     """
-    pass
+    for row in table: 
+        for element in row: 
+            if element != ".":
+                if element not in "123456789":
+                    return False 
+                else: 
+                    if row.count(element) >1: 
+                        return False 
+                    if [row[i] for i in range(9)].count(element)>1:
+                        return False 
+                    if [table[i][j] for i in range(3) for j in range(3)].count(element) >1: 
+                        return False 
+    return True 
+
+