From bc37469895c94f6c6ecaf024d1f11214594ab71b Mon Sep 17 00:00:00 2001
From: Parseluni <deliagaiaglow@gmail.com>
Date: Mon, 22 Nov 2021 08:16:00 -0800
Subject: [PATCH] completed

---
 README.md                      |  2 +-
 hash_practice/exercises.py     | 77 ++++++++++++++++++++++++++++++----
 tests/test_grouped_anagrams.py |  2 +-
 3 files changed, 70 insertions(+), 11 deletions(-)

diff --git a/README.md b/README.md
index ac48c5a..26bdf3c 100644
--- a/README.md
+++ b/README.md
@@ -13,7 +13,7 @@ By the end of this exercise you should be able to:
 
 Given an array of strings, group anagrams together.
 
-### Example
+### Example 
 
 ```
 Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..03d979d 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -1,29 +1,88 @@
 
 def grouped_anagrams(strings):
+    """
+    1. Map each ordered string as a key, appending subsequent matching ordered strings as lists
+    2. Return the list values in an array 
+    """
     """ 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) in worst case (no anagrams)
     """
-    pass
+
+    if strings == []:
+        return []
+
+    anagrams_dict = {}
+    for word in strings:                            # O(n)
+        temp = "".join(sorted(word))                # * (O(m) + O(m log(m))) = O(n * m * log(m))
+        if not anagrams_dict.get(temp): 
+            anagrams_dict[temp] = [word] 
+        else:
+            anagrams_dict[temp].append(word)
+    
+    return list(anagrams_dict.values())             # O(n) for worst case -> 2n
 
 def top_k_frequent_elements(nums, k):
+    """
+    1. Count occurrences for each num, map them in dict
+    2. Sort nums/values by value, in descending order
+    3. Return first k nums from sorted tuple list
+    """
     """ 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)
+        Space Complexity: O(n) for worst case (each element once)
     """
-    pass
+    if nums == []:
+        return []
+
+    nums_values = {}
+    for num in nums:                                # O(n)
+        if nums_values.get(num):
+            nums_values[num] += 1
+        else:
+            nums_values[num] = 1
+    
+    sorted_by_value = sorted(nums_values.items(), key = lambda x: x[1], reverse=True)
+                                                    # sorted() is Timsort -> O(n log n)
 
+    result = []
+    for i in range(k):                              # O(k) -> close to constant
+        result.append(sorted_by_value[i][0])
+    return result
 
 def valid_sudoku(table):
+    """
+    1. Check each row for duplicates
+    2. Check each column for duplicates
+    3. Check each sub grid for duplicates
+    """
     """ This method will return the true if the table is still
         a 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(1)
+        Space Complexity: O(1)
     """
-    pass
+    rows = [[] for x in range(9)]
+    columns = [[] for x in range(9)]
+    sub_grids = [[[] for x in range(3)] for x in range(3)]
+
+    for row in range(0, 9):                         # O(n)
+        for col in range(0, 9):                     # * O(n) -> O(n^2) = 81, which is the given n, which is a constant
+            if table[row][col] == ".":
+                continue
+            elif table[row][col] in rows[row]:
+                return False
+            elif table[row][col] in columns[col]:
+                return False
+            elif table[row][col] in sub_grids[row // 3][col // 3]:
+                return False
+            else:
+                rows[row].append(table[row][col])
+                columns[col].append(table[row][col])
+                sub_grids[row // 3][col // 3].append(table[row][col])
 
+    return True
\ No newline at end of file
diff --git a/tests/test_grouped_anagrams.py b/tests/test_grouped_anagrams.py
index f8e6e74..520b0f9 100644
--- a/tests/test_grouped_anagrams.py
+++ b/tests/test_grouped_anagrams.py
@@ -69,7 +69,7 @@ def test_will_work_for_strings_which_are_all_not_anagrams():
         ["pop"],
         ["pan"],
         ["pap"]
-      ];
+      ]
 
     # Assert
     assert len(answer) == 6