Hash Practice CS FUN

hash_practice/exercises.py

@@ -1,29 +1,123 @@
-
 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
+    if not strings:
+        return []
+    res_list = [[strings[0]]]
+    dict_list = [create_dict(strings[0])]
+
+    for i in range(1, len(strings)):
+        cur_str = strings[i]
+        cur_str_dict = create_dict(cur_str)
+        if cur_str_dict in dict_list:
+            match_index = dict_list.index(cur_str_dict)
+            res_list[match_index].append(cur_str)
+        else:
+            res_list.append([cur_str])
+            dict_list.append(cur_str_dict)
+    return res_list
+
+
+
+def create_dict(word): 
+    word_dict = {}
+    for letter in word:
+        if letter in word_dict:
+            word_dict[letter] += 1
+        else:
+            word_dict[letter] = 1
+    return word_dict
+
 
 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(nlogn)
+        Space Complexity: O(n)
     """
-    pass
+
+    nums_dict = create_dict(nums) 
+    num_freq_tuple_list = []
+    for num, freq in nums_dict.items():
+        num_freq_tuple_list.append((num, freq))
+    num_freq_tuple_list.sort(key=sort_by_tuple_freq, reverse=True)
+    res_list = []
+    i = 0
+    while k > 0 and i < len(num_freq_tuple_list):
+        res_list.append(num_freq_tuple_list[i][0])
+        i += 1
+        k -= 1
+    return res_list
+
 
+def sort_by_tuple_freq(x) :
+    return x[1]   
+
 
 def valid_sudoku(table):
     """ 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: ?
-    """
-    pass
+        Time Complexity: O(1)
+        Space Complexity: O(1)
+"""
+    for row in range(len(table)):
+        if valid_row(row, table) == False:   
+            return False
+
+
+    for col in range(len(table[0])):
+        if valid_col(col, table) == False:
+            return False
+
+    row = 0
+    while row < 9:
+        col = 0
+        while col < 9:
+            if valid_grid(row, col, table) == False:
+                return False
+            col += 3
+        row += 3   
+    return True
+
+
+def valid_grid(start_row, start_col, table):
+    num_set = set()
+    for i in range(start_row, start_row + 3):
+        for j in range(start_col, start_col + 3):
+            digit = table[i][j]
+            if check_valid(digit, num_set) is False:
+                return False
+    return True
+
+
+
+def valid_row(cur_row, table):
+    num_set = set()
+    for digit in table[cur_row]:
+        if check_valid(digit, num_set) is False:
+            return False
+    return True
+
+def valid_col(cur_col, table):
+    num_set = set()
+    for row in range(len(table)):
+        digit = table[row][cur_col]
+        if check_valid(digit, num_set) is False:
+            return False
+    return True
+
+
+def check_valid(digit, num_set):     
+    if digit != ".":
+        if digit in num_set:
+            return False
+        else:
+            num_set.add(digit)
+    return True