Merge pull request #1712 from ivanpenaloza/july27

adding updates

Author: ivanpenaloza

src/my_project/interviews/amazon_high_frequency_23/round_7/design_sql.py

@@ -0,0 +1,102 @@
+from typing import List, Union, Collection, Mapping, Optional, Dict
+
+class Table:
+
+    def __init__(self, name:str, columns: int, rows: Dict, row_count: int = 0):
+        self.name = name 
+        self.columns = columns
+        self.rows = rows 
+        self.row_count = row_count
+
+class SQL:
+
+    def __init__(self, names: List[str], columns: List[int]):
+        """
+        two string arrays, names and columns, both of size n.
+
+        The ith table is represented by the name names[i] and contains columns[i] number of columns
+        """
+        self.names = names 
+        self.columns = columns
+        self.data = {}
+
+        for i in range(0, len(self.names)):
+            self.data[self.names[i]] = Table(
+                name = self.names[i],
+                columns=self.columns[i],
+                rows={}
+            )
+
+        return 
+        
+
+    def ins(self, name: str, row: List[str]) -> bool:
+        """
+        Inserts row into the table name and returns true.
+        If row.length does not match the expected number of columns, or name is not a valid table,
+        returns false without any insertion.
+        """        
+        table: Table = self.data.get(name)
+        if table is None:
+            return False 
+        if len(row) != table.columns:
+            return False 
+        table.row_count += 1
+        table.rows[table.row_count] = row
+
+        return True 
+        
+
+    def rmv(self, name: str, rowId: int) -> None:
+        """
+        Removes the row rowId from the table name.
+        If name is not a valid table or there is no row with id rowId, no removal is performed.
+        """        
+        table: Table = self.data.get(name, None)
+        if table is None:
+            return 
+        
+        if rowId in table.rows:
+            del table.rows[rowId]
+
+        return 
+        
+
+    def sel(self, name: str, rowId: int, columnId: int) -> str:
+        """
+        Returns the value of the cell at the specified rowId and columnId in the table name.
+        If name is not a valid table, or the cell (rowId, columnId) is invalid, returns "<null>".
+        """        
+        null = "<null>"
+        table: Table = self.data.get(name, None)
+        if table is None:
+            return null
+        
+        if rowId not in table.rows:
+            return null
+        
+        try:
+            return table.rows[rowId][columnId - 1]
+        except Exception:
+            return null
+        
+
+    def exp(self, name: str) -> List[str]:
+        """
+        Returns the rows present in the table name.
+        If name is not a valid table, returns an empty array.
+        Each row is represented as a string, with each cell value (including the row's id) separated by a ",".
+        """
+        table: Table = self.data.get(name, None)
+        if table is None:
+            return []
+        
+        result = []
+
+        for row_id, row in table.rows.items():
+            row_str = ",".join(row)
+            result.append(f"{row_id},{row_str}")
+        
+        return result
+        
+        

src/my_project/interviews/amazon_high_frequency_23/round_7/find_minimum_time_to_finsh_all_jobs_ii.py

@@ -0,0 +1,12 @@
+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+import math
+
+class Solution:
+    def minimumTime(self, jobs: List[int], workers: List[int]) -> int:
+
+        jobs.sort()
+        workers.sort()
+
+        return max([math.ceil((n/d)) for n, d in zip(jobs, workers)])
+

src/my_project/interviews/google_top_exercises/round_1/25_word_ladder.py

@@ -3,52 +3,62 @@
 
 
 
-
 class Solution:
     def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
-        """BFS over a generic-pattern graph.
-
-        Build buckets keyed by patterns like 'h*t' so every word that differs
-        by a single letter shares a bucket. BFS from beginWord finds the
-        shortest transformation, counting words in the sequence.
         """
-
+        Find the shortest transformation sequence from beginWord to endWord.
+        Uses BFS to find the shortest path.
+        
+        Time Complexity: O(M^2 * N) where M is word length, N is wordList size
+        Space Complexity: O(M^2 * N) for the pattern dictionary
+        """
+        # If beginWord equals endWord, the sequence is just the word itself
+        if beginWord == endWord:
+            return 1
+        
+        # If endWord is not in wordList, no valid transformation exists
         if endWord not in wordList:
             return 0
-
-        # Map each wildcard pattern -> list of words matching it.
-        patterns = defaultdict(list)
-        for word in wordList:
-            for i in range(len(word)):
-                pattern = word[:i] + "*" + word[i + 1:]
-                patterns[pattern].append(word)
-
-        # BFS. Level = number of words in the sequence so far (beginWord counts as 1).
-        queue = deque([(beginWord, 1)])
+        
+        # Convert wordList to set for O(1) lookup
+        word_set = set(wordList)
+        
+        # Build a pattern dictionary to find all words that differ by one letter
+        # e.g., "hot" -> {"*ot": ["hot"], "h*t": ["hot"], "ho*": ["hot"]}
+        pattern_dict = defaultdict(list)
+        word_len = len(beginWord)
+        
+        # Add beginWord to the set if not present
+        if beginWord not in word_set:
+            word_set.add(beginWord)
+        
+        # Create patterns for all words
+        for word in word_set:
+            for i in range(word_len):
+                pattern = word[:i] + '*' + word[i+1:]
+                pattern_dict[pattern].append(word)
+        
+        # BFS to find shortest path
+        queue = deque([(beginWord, 1)])  # (current_word, level)
         visited = {beginWord}
-
+        
         while queue:
-            word, level = queue.popleft()
-
-            if word == endWord:
-                return level
-
-            for i in range(len(word)):
-                pattern = word[:i] + "*" + word[i + 1:]
-                for neighbor in patterns[pattern]:
-                    if neighbor not in visited:
-                        visited.add(neighbor)
-                        queue.append((neighbor, level + 1))
-                print(pattern,visited)
-                # Clear the bucket so it is not scanned again by another word.
-                patterns[pattern] = []
-
+            current_word, level = queue.popleft()
+            
+            # Try all possible transformations by replacing each character
+            for i in range(word_len):
+                pattern = current_word[:i] + '*' + current_word[i+1:]
+                
+                # Get all words matching this pattern
+                for next_word in pattern_dict[pattern]:
+                    if next_word == endWord:
+                        return level + 1
+                    
+                    if next_word not in visited:
+                        visited.add(next_word)
+                        queue.append((next_word, level + 1))
+                
+                # Clear the pattern to avoid revisiting in future iterations
+                pattern_dict[pattern] = []
+        
         return 0
-
-print('hello world')
-solution = Solution()
-
-print(solution.ladderLength(beginWord='hat',endWord='hut', wordList=['het',
-                                                                     'hit',
-                                                                     'hot',
-                                                                     'hut']))