From 4180d39b7de5782794e9d22f1d084e4be6b2fc18 Mon Sep 17 00:00:00 2001
From: Aida <aida.rysbek@gmail.com>
Date: Thu, 27 Jan 2022 23:19:04 -0800
Subject: [PATCH] Add working solutions to hash exercises

---
 hash_practice/exercises.py | 68 +++++++++++++++++++++++++++++++++-----
 1 file changed, 60 insertions(+), 8 deletions(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..bf04a69 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -2,18 +2,46 @@
 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(n)
     """
-    pass
+    basket = {} 
+
+    for value in strings: 
+        key = ''.join(sorted(value)) 
+        if key in basket.keys(): 
+            basket[key].append(value) 
+        else: 
+            basket[key] = [] 
+            basket[key].append(value) 
+
+    result = []
+    for key,value in basket.items(): 
+        result.append(value)
+    return result 
+
 
 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: ?
+        Space Complexity: O(n)
     """
-    pass
+    freq_map = {}
+    if nums == []:
+        return nums
+    for i in nums:
+        if i in freq_map:
+            freq_map[i] += 1
+        elif i not in freq_map:
+            freq_map[i] = 1
+    
+    result = []
+    for i in range(k):
+        highest_value = max(freq_map, key=freq_map.get)
+        result.append(highest_value)
+        freq_map.pop(highest_value)
+    return result
 
 
 def valid_sudoku(table):
@@ -22,8 +50,32 @@ 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 i in range(9):
+        board = {}
+        for j in range(9):
+            if table[i][j] in board:
+                return False
+            elif table[i][j] is not '.':
+                board[table[i][j]]
+                return True
+
+    # Check columns
+    for j in range(9):
+        for i in range(9):
+            if table[i][j] in board:
+                return False
+            elif table[i][j] is not '.':
+                board[table[i][j]]
+                return True
 
+    for subI in range(0, 9, 3):
+        for subJ in range(0, 9, 3):
+            if table[i][j] in board:
+                return False
+            elif table[i][j] is not '.':
+                board[table[i][j]]
+                return True
+    return True
\ No newline at end of file