Maple: Rhyannon Rodriguez

hash_practice/exercises.py

@@ -1,19 +1,103 @@
+import collections
+import heapq
 
 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: While there appears only a single for-loop, 
+        .join() adds O(N) & sorted() adds O(nlogn) during their ideal case
+        Space Complexity: We're adding a new data structure, the anagrams_hash,
+        meaning we're adding linear O(N) space based on the size of this new hash table
+        We're also adding an auxiliary sorted_anagram string
     """
-    pass
+    anagrams_hash = {}
+    for anagram in strings:
+        sorted_anagram = "".join(sorted(anagram))
+        if sorted_anagram in anagrams_hash:
+            anagrams_hash[sorted_anagram].append(anagram)
+        else:
+            anagrams_hash[sorted_anagram] = [anagram]
+
+    return list(anagrams_hash.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: Traversing through the list once to calculate the frequency, which is Linear O(N). 
+           In addition, we're using heap to sort & find frequent words, which is Linear-Log O(NlogK)
+           where N is the number of items in the list -> overall time complexity O(NlogK)
+        Space Complexity: Linear O(N) - we're initializing new data structures
     """
-    pass
+    if nums == []:
+        return []
+
+    nums_dict = collections.defaultdict(int)
+    for n in nums:
+        nums_dict[n] += 1
+
+    output = []
+    heap_list = [(-freq, key) for key, freq in nums_dict.items()]
+    heapq.heapify(heap_list)
+
+    while(len(output) < k):
+        output.append(heapq.heappop(heap_list)[1])
+
+    return output
+
+
+# 4 helper functions for sudoku
+def not_in_row(arr, row):
+    # Set to store characters seen so far
+    temp_set = set()
+
+    for i in range(0, 9):
+
+        # If already encountered before, return false
+        if arr[row][i] in temp_set:
+            return False
+
+        # If it is not an empty cell, insert value
+        # at the current cell in the set
+        if arr[row][i] != '.':
+            temp_set.add(arr[row][i])
+
+    return True
+
+def not_in_col(arr, col):
+    temp_set = set()
+    for i in range(0, 9):
+        # If already encountered before,return false
+        if arr[i][col] in temp_set:
+            return False
+
+        # If it is not an empty cell, insert
+        # value at the current cell in the set
+        if arr[i][col] != '.':
+            temp_set.add(arr[i][col])
+
+    return True
+
+def not_in_box(arr, startRow, startCol):
+    temp_set = set()
+    for row in range(0, 3):
+        for col in range(0, 3):
+            curr = arr[row + startRow][col + startCol]
+
+            # If already encountered before, return false
+            if curr in temp_set:
+                return False
+
+            # If it is not an empty cell,
+            # insert value at current cell in set
+            if curr != '.':
+                temp_set.add(curr)
+
+    return True
+
+def is_valid(arr, row, col):
+    return (not_in_row(arr, row) and not_in_col(arr, col) and
+            not_in_box(arr, row - row % 3, col - col % 3))
 
 
 def valid_sudoku(table):
@@ -22,8 +106,17 @@ 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: ?
-    """
-    pass
+        Time Complexity: Quadratic O(N^2) - nested for loop
+        Space Complexity: Constant O(1) - not initializing any new data structures;
+        unless we're counting temp sets from our helper functions
 
+    """
+    length = len(table)
+    for i in range(0, length):
+        for j in range(0, length):
+            # If current row or current column or
+            # current 3x3 box is not valid, return false
+            if not is_valid(table, i, j):
+                return False
+
+    return True